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>
#include<stdlib.h>
int main()
{
  int A[3][3],N,b[10],i,j,k,num,num1,l,m,n,p,q;
  int check[3][3]={};
  for(i=0;i<3;i++)
  {
    for(j=0;j<3;j++)
    {
      scanf("%d",&A[i][j]);
    }
  }
  scanf("%d",&N);
  for(k=0;k<N;k++)
  {
    scanf("%d",&b[k]);
  }
  num=0;
  num1=0;
  for(l=0;l<N;l++)
  {
    for(m=0;m<3;m++)
    {
      for(n=0;n<3;n++)
      {
        if(b[l]==A[m][n])
        {
          check[m][n]=1;
        }
      }
    }
  }
 
  
  for(p=0;p<3;p++)
  {
    for(q=0;q<3;q++)
    {
     if(check[p][q]==1)
     {
       num=num+1;
     }
     if(num==3)
     {
      printf("Yes");
       exit(0);
     }
     if(check[q][p]==1)
     {
       num1=num1+1;
     }
     if(num1==3)
     {
      printf("Yes");
       exit(0);
     }
    }
  }
  if(check[0][0]==1&&check[1][1]==1&&check[2][2]==1)
  {
    printf("Yes");
    exit(0);
  }
  if(check[2][0]==1&&check[1][1]==1&&check[0][2]==1)
  {
    printf("Yes");
    exit(0);
  }
  printf("No");
}