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(void){
  int i,j,n,t;
  int s;
  int m;
  int p=0;
  int q=0;
  int a[3][3];
  int b[100];
  int c[3][3];
  for(i=0;i<3;i++)
    for(j=0;j<3;j++)
      scanf("%d",&a[i][j]);
  scanf("%d",&n);
  for(t=0;t<n;t++){
    scanf("%d",&b[t]);
    for(i=0;i<3;i++)
      for(j=0;j<3;j++)
	if(b[t]==a[i][j]){
	  c[i][j]=1;
	}else{
	  c[i][j]=0;
	}
  }

  for(i=0;i<3;i++){
    s=0;
    for(j=0;j<3;j++){
      s+=c[i][j];
    }
    if(s==3){
      printf("Yes");
      return 0;
    }
  }

  for(j=0;j<3;j++){
    m=0;
    for(i=0;i<3;i++){
      m+=c[i][j];
    }
    if(m==3){
      printf("Yes");
      return 0;
    }
  }

   for(i=0;i<3;i++){
      p+=c[i][i];
    }
    if(p==3){
      printf("Yes");
      return 0;
    }

    for(i=0;i<3;i++){
      q+=c[i][2-i];
    }
    if(q==3){
      printf("Yes");
      return 0;
    }

    printf("No");
    return 0;

}