Problem:
Score : 200 points

Problem Statement

We have a bingo card with a 3\times3 grid. The square at the i-th row from the top and the j-th column from the left contains the number A_{i, j}.

The MC will choose N numbers, b_1, b_2, \cdots, b_N. If our bingo sheet contains some of those numbers, we will mark them on our sheet.

Determine whether we will have a bingo when the N numbers are chosen, that is, the sheet will contain three marked numbers in a row, column, or diagonal.

Constraints

All values in input are integers.

1 \leq A_{i, j} \leq 100

A_{i_1, j_1} \neq A_{i_2, j_2} ((i_1, j_1) \neq (i_2, j_2))

1 \leq N \leq 10

1 \leq b_i \leq 100

b_i \neq b_j (i \neq j)

Input

Input is given from Standard Input in the following format:

A_{1, 1} A_{1, 2} A_{1, 3}
A_{2, 1} A_{2, 2} A_{2, 3}
A_{3, 1} A_{3, 2} A_{3, 3}
N
b_1
\vdots
b_N

Output

If we will have a bingo, print Yes; otherwise, print No.

Sample Input 1

84 97 66
79 89 11
61 59 7
7
89
7
87
79
24
84
30

Sample Output 1

Yes

We will mark A_{1, 1}, A_{2, 1}, A_{2, 2}, A_{3, 3}, and complete the diagonal from the top-left to the bottom-right.

Sample Input 2

41 7 46
26 89 2
78 92 8
5
6
45
16
57
17

Sample Output 2

No

We will mark nothing.

Sample Input 3

60 88 34
92 41 43
65 73 48
10
60
43
88
11
48
73
65
41
92
34

Sample Output 3

Yes

We will mark all the squares.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int main(){
   int a[3][3];
   int i,j,n,b,h,judge = 0,ans = 0;
   for(i = 0; i < 3; i++){
      for(j = 0; j < 3; j++){
         scanf("%d",&a[i][j]);
      }
   }
   scanf("%d",&n);
   for(h = 0; h < n; h++){
      judge = 0;
      scanf("%d",&b);
      for(i = 0; i < 3; i++){
         for(j = 0; j < 3; j++){
            if(b==a[i][j]){
               a[i][j]=0;
               judge = 1;
               break;
            }
         }
         if(judge == 1){
            break;
         }
      }
   }

   
   if(a[0][0] == 0){
      if(a[0][1] == 0 && a[0][2] == 0){
         ans = 1;
      }else if(a[1][0] == 0 && a[2][0] == 0){
         ans = 1;
      }else if(a[2][2] == 0 && a[1][1] == 0){
         ans = 1;
      }
   }

   if(a[0][2] == 0){
      if(a[1][2] == 0 && a[2][2] == 0){
         ans = 1;
      }else if(a[1][1] == 0 && a[2][0] == 0)
         ans = 1;
   }

   if(a[0][1] == 0 && a[1][1] == 0 && a[2][1] == 0){
      ans = 1;
   }

   if(a[1][0] == 0 && a[1][1] == 0 && a[1][2] == 0){
      ans = 1;
   }

   if(a[2][0] == 0 && a[2][1] == 0 && a[2][2] == 0){
      ans = 1;
   }

   if(ans == 1){
      printf("Yes");
   }else{
      printf("No");
   }
   return 0;
}