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 res=0;
    int row=0;
    int column=0;
    int i,j,N,k=0;
    int arr[3][3];
    char result[3][3];
    int* arr_;
    
    for(i=0;i<3;i++)
    {
        scanf("%d%d%d",&arr[i][0],&arr[i][1],&arr[i][2]);
    }
    scanf("%d",&N);
    arr_ = (int*)malloc(N*sizeof(int));
    for(i=0;i<N;i++)
    {
        scanf("%d",&arr_[i]);
    }
    
    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            for(k=0;k<N;k++)
            {
                if(arr_[k]==arr[i][j])
                {
                    result[j][i]= 'x';
                    break;
                }
                else
                {
                    result[j][i]= '*';
                }
            }
        }
    }
    
    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            if(result[i][j] == 'x')
            {
                row++;
            }
            for(k=0;k<3;k++)
            {
                if(result[k][j] == 'x')
                {
                    column++;
                }   
            }
            if(column==3)
            {
                res=1;
                break;
            }
            else
            {
                 column=0;
            }
        }
        if(row==3)
        {
            res=1;
            break;
        }
        else
        {
            row=0;
        }
    }
        
    if(result[1][1]='x')
    {
        if(result[0][0]== 'x' && result[2][2] == 'x')
        {
            res=1;
        }
        if(result[0][2]== 'x'  && result[2][0] == 'x')
        {
            res=1;
        }
    }
    
    if(res==1)
    {
        printf("Yes");
    }
    else
    {
        printf("No");
    }
    
    return 0;
}