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: Runtime Error)

#include <stdio.h>

char *main()
{
  int i;
  int j;
  int sum;
  int a[3][3] = {0};
  int check[3][3] = {0};
  int b;
  
  i = -1;
  while (++i < 3)
    scanf("%d %d %d",&a[i][0],&a[i][1],&a[i][2]);

  while (scanf("%d", &b) != EOF)
  {
	i = -1;
    while (++i < 3)
    {
      j = -1;
      while (++j < 3)
      {
    	if (a[i][j] == b)
          check[i][j] = 1;
      }
    }
  }
  
  // 横方向
  i = 0;
  while (i < 3)
  {
    j = 0;
    sum = 0;
  	while (j < 3)
    {
      sum+= check[i][j];
      j++;
    }
    if (sum == 3)
      return ("Yes");
    i++;
  }
  
  // 縦方向
  j = 0;
  while (j < 3)
  {
    i = 0;
    sum = 0;
  	while (i < 3)
    {
      sum+= check[i][j];
      i++;
    }
    if (sum == 3)
      return ("Yes");
    j++;
  }
  
  // 斜め方向 右上から左下
  sum = 0;
  while (i < 3)
  {
    sum+= check[i][i];
    i++;
  }
  if (sum == 3)
    return ("Yes");
  
// 斜め方向 左上から右下  
  sum = 0;
  while (i < 3)
  {
    sum+= check[2 - i][i];
    i++;
  }
  if (sum == 3)
    return ("Yes");
  
  return ("No"); 
}