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               B j ( z1 , z2 ,... zn )U j W j − L j + 1 − π j I j − z j + 1 − B j ( z1 , z2 ,... zn ) U j (W j − π j I j − z j )




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π j = Bj

                λ                                                                   π j = (1 + λ ) B j #         D$
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        I=L                                                                                                 #,$




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        max B ( z )U (W − L − z ) + 1 − B ( z ) U (W − z )                                                  #6$
          z




                                                   U ′ (W − z )
                       B ( z ) + (1 − B ( z ) )
                                                  U ′ (W − L − z )        1
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    B1 ( z1 , z2 ) = p ( z1 ) + 1 − p ( z1 ) qp ( z2 ) = 1 − 1 − p ( z1 ) 1 − qp ( z2 )                            #0$

                                                                 /       1 − p ( z1 ) 1 − qp ( z2 )

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    π 1 ( z1 : z2 ) = [1 + λ ] B1 ( z1 , z2 )                                                                      #@$




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    max B1 ( z1 , z2 ) U j (W − L + 1 − π ( z1 : z2 ) I1 − z1 ) + 1 − B1 ( z1 , z2 ) U j (W − π ( z1 : z2 ) I1 − z1 )                #D$
     z1 , I1


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p ( z ) + 1 − p ( z ) qp ( z ) > p( z )                                                                 K

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p ( z1 ) (1 − qp ( z2 ) )                           !

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       z1 , z2 , I1 , I 2
                            j =1



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                                                  λ
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                                {
     π ( z1 , z2 ) = (1 + λ ) 2qp ( z1 ) (1 − p ( z2 ) ) + p ( z1 ) − qp ( z1 ) (1 − p ( z2 ) )                 }   !

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         max B1U j (W − L + [1 − π 1 ] I1 − z1 ) + [1 − B1 ]U j (W − π 1 I1 − z1 )                                                    #&'$
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                                               λ
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    Political Economy, Vol. 80,No. 4,1972, pg. 623-648.
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Appendix
Proof 1. Baseline scenario:
Objective based on the expected utility function of the firm is
                     max B ( z ) U B (W − L + [1 − π ( z )] I − z ) + 1 − B ( z ) U N (W − π ( z ) I − z )
                              z, I
The first order condition for IT security investment is
                                                   ∂π ( z )
                     B′ ( z ) [U B − U N ] − 1 +            I B ( z )U B + {1 − B ( z )}U N = 0 .
                                                                        ′                 ′                             (A1)
                                                      ∂z
First order condition for insurance is
                     B ( z ) [1 − π ( z )]U B − π ( z ) 1 − B ( z ) U N = 0 .
                                            ′                         ′                                                 (A2)
Since   π ( z ) = B ( z ) , from the latter condition we see that U B = U N
                                                                    ′     ′                       and I=L. That is, the marginal utilities in both
states are equal. From the first condition,
                                       1
                           B′( z ) = −                                                                              (A3)
                                       L
Proof 2. IT Security Spending with No Insurance Market
Utility function of the firm is
                                      max B ( z ) U B (W − L − z ) + (1 − B ( z ) ) U N (W − z )
                                               z
The first order condition for IT security investment is
                                                                    ′  {
                                   B′ ( z ) [U B − U N ] − B ( z )U B + (1 − B ( z ) )U N = 0
                                                                                        ′                  }
For W large enough, first order Taylor series approximation gives
                             UN ≈ UB +UBL  ′                                                                   (A4)
                                                                   {                                   }
                                                   − B ′ ( z )U B L − B ( z )U B + (1 − B ( z ) )U N = 0
                                                                ′              ′                   ′
                                                                                                   ′
                                                                                                  UN
                                                                       B ( z ) + (1 − B ( z ) )
                                                                                                   ′
                                                                                                  UB
                                                        B′ ( z ) = −
                                                                                    L
                                  ′
                                 UN
since B ( z ) + (1 − B ( z ) )      < 1 , IT security investment when insurance is available at fair market price is lower
                                  ′
                                 UB
than IT security investment when there is no insurance market available.
Proof 3. Interdependent case:
Utility function of the firm 1 is
             max B1 ( z1, z2 )U B1 (W − L + [1 − π ( z1 : z2 )] I1 − z2 ) + 1 − B ( z1 , z2 ) U N 1 (W − π ( z1 : z2 ) I1 − z2 )
                 z1 , I1

where B1 ( z1 , z2 ) = 1 − (1 − p ( z1 ) ) (1 − qp ( z2 ) )
The first order condition for IT security investment is
       ∂B1 ( z1 , z2 )                       ∂π ( z1 , z2 )
            ∂z1
                       (U B1 − U N 1 ) − 1 +
                                                ∂z1
                                                                  {
                                                            I1 B1 ( z1 , z2 )U B1 + 1 − B1 ( z1 , z2 ) U N 1 = 0
                                                                               ′                         ′      }                     (A5)

First order condition for insurance is
                               B1 ( z1 , z2 )(1 − π 1 ( z1 , z2 ))U B1 − (1 − B1 ( z1 , z2 ))π 1 ( z1 , z2 )U N 1 =0
                                                                    ′                                         ′
                             ∂B1 ( z1 , z2 )                              1
If =0, then I=L and                          = p′( z1 )(1 − qp( z2 )) = −
                                 ∂z1                                      L



                                                                            26
If >0, for W large enough, using first order Taylor series approximation
                                U N 1 ≈ U B1 + U B1 ( L − I 1 ) ; U N 1 ≈ U B1 + U B1 ( L − I 1 )
                                                 ′                  ′       ′      ′′
                       Substituting in A5, dividing by U′ 1 and using first order condition for insurance
                                                        B

                                                     U N 1 B1 ( z1 , z2 ) (1 − [1 + λ ] B1 ( z1 , z2 ) )
                                                       ′
                                                          =
                                                       ′
                                                     U B1        [1 + λ ] (1 − B1 ( z1 , z2 ) )
                                                                           we get
                                                                ∂B1 ( z1 , z2 )       −1
                                                                                =
                                                                    ∂z1           [1 + λ ] L
and assuming the CARA utility function we get from the FOC for insurance
                                                                                           λ
                                                 r [ L − I1 ] =
                                                                  [1 + λ ] 1 − p ( z1 ) 1 − qp ( z2 )
               U ′′
where r = −         is a constant and greater than 0. Using identical firms,
               U′
                                                                    1
                                      p′ ( z ) 1 − qp ( z ) = −                                                  (A6)
                                                                [1 + λ ] L
                                                                              λ
                                            I = L−                                                               (A7)
                                                       r [1 + λ ] 1 − p ( z ) 1 − qp ( z )
Proof 4: Condition for Unique Equilibrium
From the first order condition of IT security investment (A6),
                                                                                                           1
                                           Π1 ( R1 ( z2 ) , z2 ) = p′ ( z1 ) 1 − qp ( z2 ) +                      =0
                                            1
                                                                                                       [1 + λ ] L
The slope of reaction function for Firm 1 and 2 is
                      Π1 ( R1 ( z2 ) , z2 ) p′′ ( z1 ) (1 − qp ( z2 ) )                Π 2 ( R2 ( z1 ) , z1 )      p′ ( z1 ) qp′ ( z2 )
        R1′ ( z2 ) = − 1                                                ; R2 ( z1 ) = − 2
                       12                                                                21
                                             =                             ′                                  =
                      Π11 ( R1 ( z2 ) , z2 )    p′ ( z1 ) qp′( z2 )                    Π 22 ( R2 ( z1 ) , z1 ) p ′′ ( z1 )(1 − qp( z2 ) )
In order for reaction curve to intersect, the slope of                 R1 should be higher than the slope of R2 . So
                                                  p′′ ( z1 ) (1 − qp ( z2 ) )              p ′ ( z1 ) qp′ ( z2 )
                                                                                  >                               .
                                                      p ′ ( z1 ) qp ′( z2 )            p ′′ ( z1 )(1 − qp( z2 ) )
cross-multiplying and rearranging
                   {
                 p′′ ( z1 )(1 − qp( z2 ) ) − p ′ ( z1 ) qp ′ ( z2 )           }{      p′′ ( z1 )(1 − qp( z2 ) ) + p′ ( z1 ) qp′ ( z2 )   }> 0
Note that the second term in the LHS multiplicand is positive. Hence for an unique equilibrium,
                                     p′′ ( z1 )(1 − qp( z2 ) ) − p′ ( z1 ) qp′ ( z2 ) > 0
Assuming symmetric firms, the condition for unique equilibrium can be written as
                                                       p′′ ( z )(1 − qp( z ) ) − q p′ ( z )
                                                                                                       2
                                                                                                           >0
Proof 5:
Denote the level of IT security investment and insurance coverage taken in independent firm and dependent firms
as   z I and z D respectively and I I and I D .
                                     1                          λ
                 p′ ( z I ) = −            ; r L−II =                                                                              (A8(a,b))
                                  [1+ λ] L            [1 + λ ] 1 − p ( z I )
                                                1                                   λ
            p′ ( z D ) 1 − qp ( z D ) = −              ; r L−ID =                                                                         (A9(a,b))
                                            [1 + λ ] L            [1 + λ ] 1 − p ( z ) 1 − qp ( z D )
                                                                                    D


Dividing A8a and A9a,
                                      p′ ( z I ) = p ′ ( z D ) 1 − qp ( z D ) ; p′ ( z I ) > p ′ ( z D )              zI > zD
If >0, dividing A8b and A9b,


                                                                                27
                                                              L−II            1 − p ( z D ) 1 − qp ( z D )
                                                                          =
                                                              L− ID                         1− p (zI )

                                                                                        L−II
                     z >z
                        I    D
                                     1− p (z      I
                                                      )       > 1− p (z   D
                                                                              )    ;                < 1 or I I > I D (if =0, I I = I D ).
                                                                                       L−I      D


Proof 6:
                        ∂z               p ( z ) p′ ( z )
                           =                                           < 0 since denominator is less than zero.
                        ∂q p′′ ( z ) 1 − qp ( z ) − p′ ( z ) qp′ ( z )

                                                      ∂I         λ p ′ ( z ) 1 + q − 2qp ( z )
                                             (i)         =−                                                              >0
                                                      ∂z    [1 + λ ] r 1 − p ( z ) 1 − qp ( z )
                                                                                    2                                2



                        ∂I ∂z      λ p′ ( z ) 1 + q − 2qp ( z )                                              p ( z ) p′ ( z )
                              =                                                                                                                 ≤0
                        ∂z ∂q [1 + λ ] r 1 − p ( z ) 2 1 − qp ( z )                     2
                                                                                            { p′′ ( z ) 1 − qp ( z )                        }
                                                                                                                         − p′ ( z ) qp′ ( z )
                                            ∂z                             1
                                     (ii)      =                                                             >0
                                            ∂L [1 + λ ] L2 p ′′ ( z ) (1 − qp ( z ) ) − p ′ ( z ) qp ′ ( z )

           ∂I ∂I ∂z           λ p ′ ( z ) 1 + q − 2qp ( z )
             +      = 1−                                                                                                                                    >0
           ∂L ∂z ∂L      [1 + λ ] r 1 − p ( z ) 1 − qp ( z )                                [1 + λ ] L2   p′′ ( z ) (1 − qp ( z ) ) − p ′ ( z ) qp′ ( z )
                                                 2                                     2



                                    ∂z                                            ∂I                    λ
                            (iii)      =0,                                           =                                           ≥0
                                    ∂r                                            ∂r r 2 (1 + λ ) (1 − p ( z ) ) (1 − qp ( z ) )
Proof 7:
              ∂π            ∂B ∂z                                            ∂B
                 = (1 + λ )       , B ( z ) = 1 − 1 − qp ( z ) 1 − p ( z ) ,    = p′ ( z ) 1 + q − 2qp ( z ) < 0 and
              ∂q            ∂z ∂q                                            ∂z
                                    ∂z                                              ∂π
                                        < 0 from proposition 2 (i): As a result,          >0
                                    ∂q                                               ∂q
Proof 8:
                                          ∂z                            1
                                             =                                                            >0
                                                     2
                                                                      {
                                          ∂λ [1 + λ ] L p ′′ ( z ) (1 − qp ( z ) ) − p ′ ( z ) qp ′ ( z )                  }
∂I ∂I ∂z ∂I         λ p ′ ( z ) 1 + q − 2qp ( z )
  =     +   =−
∂λ ∂z ∂λ ∂λ    [1 + λ ] r 1 − p ( z ) 1 − qp ( z )
                                       2                                      2
                                                                                   [1 + λ ]2 L { p ′′ ( z ) (1 − qp ( z ) ) − p′ ( z ) qp′ ( z )}
                                                                                        1
                                                               −
                                                                   1 − p ( z ) 1 − qp ( z ) r [1 + λ ]
                                                                                                                 2



                    1                                                                  λ p′ ( z ) 1 + q − 2qp ( z )
=                                             −                                                                                                             −1
    1 − p ( z ) 1 − qp ( z ) r [1 + λ ]
                                        2
                                                  (1 + λ )                                            {
                                                                   1 − p ( z ) 1 − qp ( z ) L p ′′ ( z ) 1 − qp ( z ) − p′ ( z ) qp′ ( z )             }
                                            p ′ ( z ) 1 + q − 2qp ( z )
Denote λ * = −                                                                                                   >0
                                                          {
                   1 − p ( z ) 1 − qp ( z ) L p′′ ( z ) 1 − qp ( z ) − p′ ( z ) qp′ ( z )                    }
                    ∂I                   1                                     λλ *                  1    ∂I            ∂I
                       =                                                               − 1 . If λ > *   ,    > 0 , else    <0.
                    ∂λ   1 − p ( z ) 1 − qp ( z ) r [1 + λ ]
                                                             2
                                                                              (1 + λ )             λ − 1 ∂λ             ∂λ
Proof 9:
                            λ                                                 λ                                      λ
I = L−                                                    = L−                                   = L−
         r (1 + λ ) 1 − p ( z ) 1 − qp ( z )                       r (1 + λ ) [1 − B ( z )]               r (1 + λ − π ( z ) )




                                                                                       28
                                                                      ∂π ( z )                      ∂B ∂z
                                                                               = B ( z ) + (1 + λ )
                                                                       ∂λ                           ∂z ∂λ
                                                                              λ                                  ∂π ( z )
                                                                  ∂                                1−π (z) + λ
                                                     ∂I               (1 + λ − π ( z ) )                            ∂λ
                                                        =−                                    =−
                                                     ∂λ                      ∂λ                      (1 − λ − π ( z ) )
                                                                                                                        2



                                                                                                       ∂π ( z )     ∂I
We know that insured amount will greater than premium paid.(i.e. ( 1 − π ( z ) > 0 )As a result, if             >0,    <0.
                                                                                                         ∂λ         ∂λ
If loading factor increases, insurance coverage will decreases if price of insurance ( (z)) increases as well.
                                                                       ∂B ∂z
                      ∂π ( z )                                       −                   ∂π ( z )
                                                    ∂B ∂z
                               = B ( z ) + (1 + λ )       > 0 (if λ > ∂z ∂λ − 1 = λ ** ,          > 0 ).
                        ∂λ                          ∂z ∂λ              B (z)              ∂λ
Proof 10:
                                  −1                                         −1                               λ
In region 1; p′ ( z1 ) =             , I1 = L . In region 2; p′ ( z2 ) =            , r ( L − I2 ) =                       . In region 3;
                                  L                                      (1 + λ ) L                  (1 + λ ) 1 − p ( z2 )
                        −1                                                            −1
 p ′ ( z3 ) =                        , I 3 = L . In region 4, p′ ( z4 ) =                             ,
                  L (1 − qp ( z3 ) )                                      L (1 + λ ) (1 − qp ( z4 ) )
                                         λ
r ( L − I4 ) =
                     (1 + λ )   1 − p ( z4 ) 1 − qp ( z4 )
                                       ∂z
z1 > z3 , z2 > z4 , z2 > z1 and since      > 0 , z4 > z3 .
                                       ∂λ
As a result, z2 > z4 , z1 > z3 . For the insurance amount, since z2 > z4 , I1 = I 3 > I 2 > I 4 . Traditional Insurance Market
(q=0 , =0) versus Current cyber insurance market. We will compare IT security investment level in region 4 and in
              region 1. For cyber insurance coverage taken L = I1 > I 4 . For the IT security investment,
                                                                      p ′ ( z1 )
                                                                                 = [1 + λ ] 1 − qp ( z4 )    .
                                                                      p ′ ( z4 )
If [1 + λ ] 1 − qp ( z4 ) = 1 , then z1 = z4 .
If [1 + λ ] 1 − qp ( z4 ) < 1                  p′ ( z1 ) > p ′ ( z4 )       z1 > z4
If [1 + λ ] 1 − qp ( z4 ) > 1                  p′ ( z1 ) < p ′ ( z4 )       z1 < z4
Proof 11: Joint decision-making solution
Suppose that there are two firms; firm 1 and firm 2. Social planner will maximize the following
B1U B1 (W − L + (1 − π 1 ) I1 − z1 ) + (1 − B1 )U N 1 (W − π 1 I1 − z1 ) + B2U B 2 (W − L + (1 − π 2 ) I 2 − z2 ) + (1 − B2 )U N 2 (W − π 2 I 2 − z2 )
where B1 = 1 − (1 − p ( z1 ) ) (1 − qp ( z2 ) ) , B2 = 1 − (1 − p ( z2 ) ) (1 − qp ( z1 ) ) and π 1 = (1 + λ ) B1
FOC for self protection with respect to                       z1 ;
∂B1                                                               ∂π 1     ∂B                                                    ∂π 2
    (U B1 − U N 1 ) − B1U B1 + (1 − B1 )U N 1
                          ′               ′                  1+        I1 + 2 (U B 2 − U N 2 ) − B2U B 2 + (1 − B2 )U N 2
                                                                                                     ′                ′               I2 = 0
∂z1                                                               ∂z1      ∂z1                                                   ∂z1
FOC for insurance is;
                                                                  B1 (1 − π 1 )U B1 − (1 − B1 ) π 1U N 1 = 0
                                                                                 ′                   ′
             st
Taylor 1 order approximation yields as before
∂B1                      ′
                       U N1     ∂π 1     ∂B2                                              ′
                                                                                        U B2     U′              U′           ∂π 2
       ( I1 − L ) −    B1 + (1 − B1 )            1+          I1 +           ( I2 − L)        − B1 B 2 + (1 − B1 ) N 2              I2 = 0
 ∂z1                                      ′
                                        U B1           ∂z1            ∂z1                 ′
                                                                                        U B1       ′
                                                                                                 U B1              ′
                                                                                                                 U B1         ∂z1

For identical agent, U B1 = U B 2 = U B and U N 1 = U N 2 = U N and since
                                                                                                          (1 − (1 + λ ) B ) = U N
                                                                                                                                ′
                                                                                                          (1 − B ) [1 + λ ]     ′
                                                                                                                               UB



                                                                                         29
                                                                                                   1
                                                           p′ ( z ) (1 + q − 2qp ( z ) ) = −
                                                                                                (1+ λ ) L
                                                                                                1
The above equation can be written as p′ ( z ) (1 − qp ( z ) ) + M = −                                  ; M ′ = p′ ( z ) ( q − qp ( z ) ) ≤ 0
                                                                                            (1 + λ ) L
Comparing with individual firm’s first order condition
                                     1                           ∂z
 p′ ( z ) (1 − qp ( z ) ) + M = −           where M=0 and since     <0.
                                  (1+ λ ) L                     ∂M
Proof 12:
Utility of firm 1 will be
qp ( z1 ) (1 − p ( z2 ) )U A (W − 2 L + 2 I1 − π ( z1 , z2 ) I1 − z1 ) +
  p ( z1 ) − qp ( z1 ) (1 − p ( z2 ) ) U B (W − L + I1 − π ( z1 , z2 ) I1 − z1 ) + (1 − p ( z1 ) )U C (W − π ( z1 , z2 ) I1 − z1 )
First order condition with respect to z1
qp ′ ( z1 ) (1 − p ( z2 ) )U A + p′ ( z1 ) − qp′ ( z1 ) (1 − p ( z2 ) ) U B − p′ ( z1 )U C
       ∂π ( z1 , z2 )
−(1 +                   I1 ) qp ( z1 ) (1 − p ( z2 ) ) U A + p ( z1 ) − qp ( z1 ) (1 − p ( z2 ) ) U B + (1 − p( z1 ) )U C = 0
                                                         ′                                            ′                   ′
             ∂z1
First order condition with respect to I1
( 2 − π ( z1 , z2 ) ) qp ( z1 ) (1 − p ( z2 ) )U A + (1 − π ( z1 , z2 ) ) p ( z1 ) − qp ( z1 ) (1 − p ( z2 ) ) U B − (1 − p( z1 ) ) π ( z1 , z2 )U C = 0
                                                 ′                                                               ′                                 ′
Following first order Taylor series approximation,
              ′                            ′                 ′     ′     ′′
U B ≈ U A + U A ( L − I ) , U C ≈ U A + 2U A ( L − I ) and U B ≈ U A + U A ( L − I ) .
From first order condition with respect to I1 and replacing (1 − p( z1 ) )U C in first order condition,
                                                                            ′

  p′ ( z1 ) − qp′ ( z1 ) (1 − p ( z2 ) ) U A ( L − I ) − p′ ( z1 ) 2U ′ ( L − I )
                                           ′                          A

        ∂π ( z1 , z2 )      2qp ( z1 ) (1 − p ( z2 ) )      p ( z1 ) − qp ( z1 ) (1 − p ( z2 ) )
−(1 +                  I1 )                             ′
                                                       UA +                                       ′
                                                                                                 UB = 0
           ∂z1                   π ( z1 , z2 )                         π ( z1 , z2 )

Substituting the Taylor approximation, dividing by                         U ′ , and since , and for symmetric firms,
                                                                             A

                                                                                               ∂π ( z , z )    p ( z ) − qp ( z ) (1 − p ( z ) )
                                                                                        (1 +                I)                                   [ r( L − I )]
                                                       1                                          ∂z                      π ( z, z )
   p′ ( z ) + qp′ ( z ) − qp′ ( z ) p ( z ) = −               + K , where K =                                                                                    >0
                                                   (1 + λ ) L                                                                L
                                                                                                     −1
The equation above can be written as                    p′ ( z ) − qp′ ( z ) p ( z ) − K ′ =                 , where K ' = K − qp′ ( z ) > 0
                                                                                                [1 + λ ] L
                                                                                                         ∂z                      1
For the individual choice of z earlier               K ' =0.From the previous equation                      =                                            >0
                                                                                                        ∂K ′ p′′ ( z ) 1 − qp ( z ) − qp′ ( z ) p′ ( z )
As a result IT security investment with liability is greater than without liability. The equation above can be written as
                                        −1
p′ ( z ) 1 + q − 2qp ( z ) − K ′′ =            where K ′′ = K − qp′ ( z ) p ( z ) > 0 .
                                    [1 + λ ] L
For the joint choice of z earlier, K′′ =0. Thus, from the last equation
                                     ∂z                          1
                                         =                                                  >0
                                    ∂K ′′ p′′ ( z ) 1 + q − 2qp ( z ) − 2qp′ ( z ) p′ ( z )
IT security investment level with liability is higher than social optimum level of IT security investment without
liability.
Proof 13: Generalization to Several Interdependent Firms
Proof is omitted due to space limitation. However, proof is available from authors upon request.




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