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: Time Limit Exceeded)

#include <stdio.h>

int main(void)
{
        int a[3][3],n,i,j,k;

        for(i=0;i<3;i++){
                for(j=0;i<3;j++)
                        scanf("%d",&a[i][j]);
        }
        scanf("%d",&n);
        int b;

        for(i=0;i<n;i++){
                scanf("%d",&b);
                for(j=0;j<3;j++)
                        for(k=0;k<3;k++)
                                if(a[j][k]==b){ a[j][k]=0; break;}
        }
        for(i=0;i<3;i++){
                for(j=0;j<3;j++){
                        printf("%d ",a[i][j]);
                }
                printf("\n");
        }

        for(i=0;i<3;i++)
                if(a[i][0]+a[i][1]+a[i][2]==0) {printf("Yes\n"); return 0;}
        for(i=0;i<3;i++)
                if(a[0][i]+a[1][i]+a[2][i]==0){printf("Yes\n"); return 0;}

        if(a[0][0]+a[1][1]+a[2][2]==0){ printf("Yes\n"); return 0;}
        if(a[0][2]+a[1][1]+a[2][0]==0){ printf("Yes\n"); return 0;}

        printf("No\n");
        return 0;
}