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 main()
{
  int N[3][3];
  int B[3][3];
  int nb, n;
  int i, j, k;
  i = 0;
  while (i < 3)
  {
    scanf("%d %d %d", &N[i][0], &N[i][1], &N[i][2]);
    i++;
  }
  scanf("%d", &nb);
  i = 0;
  while (i < nb)
  {
    scanf("%d", &n);
    j = 0;
    while (j < 3)
    {
      k = 0;
      while (k < 3)
      {
        if (n == N[j][k])
        {
          B[j][k] = 1;
        }
        k++;
      }
      j++;
    }
    i++;
  }
  //行
  i = 0;
  while (i < 3)
  {
    if (B[i][0] + B[i][1] + B[i][2] == 3)
    {
      printf("Yes\n");
      return (0);
    }
    i++;
  }
  //列
  i = 0;
  while (i < 3)
  {
    if (B[0][i] + B[1][i] + B[0][i] == 3)
    {
      printf("Yes\n");
      return (0);
    }
    i++;
  }
  //斜め
  if (B[0][0] + B[1][1] + B[2][2] == 3)
    {
      printf("Yes\n");
      return (0);
    }
  if (B[2][0] + B[1][1] + B[0][2] == 3)
    {
      printf("Yes\n");
      return (0);
    }
  printf("No\n");
}