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: Wrong Answer)

#include <stdio.h>
int b[3][4];
int main (){
  int bingo[3][4],n, a, c=0;

  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++)
      scanf("%d", &bingo[i][j]);
  }

  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);

    for(int k = 0; k < 3; k++)
    {
      for (int j = 0; j < 3; j++)
       {
        if (bingo[k][j] == a)
        {
          c++;
          b[k][j]++;
          break;
        }
      }
    }
  }

  if (c>=3){
    for (int i = 0; i < 3; i++) {
        if (b[i][0]==1 && b[i][1]==1 && b[i][2]==1){
           printf("Yes\n");
           return 0;
         }
        if(b[0][i]==1 && b[1][i]==1 && b[2][i]==1){
          printf("Yes\n");
          return 0;
        }
       }
        if( b[0][0]==1 && b[1][1]==1 && b[2][2]==1){
          printf("Yes\n");
          return 0;
        }
        if( b[0][2]==1 && b[1][1]==1 && b[2][0]==1){
          printf("Yes\n");
          return 0;
        }
  }
  else
    printf("No\n");

  return 0;
}