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: Accepted)

#include <stdio.h>
#include <stdlib.h>

int main(void){
    int a[3][3];
    int chk[9] ={0};
    int n;
    int ii1, ii2, ii3;
    for(ii1=0; ii1<3; ii1++){
        scanf("%d %d %d", &a[ii1][0], &a[ii1][1], &a[ii1][2]);
    }
    scanf("%d", &n);
    int b[n];
    for(ii1=0; ii1<n; ii1++){
        scanf("%d", &b[ii1]);
    }

    for(ii1=0; ii1<n; ii1++){
        for(ii2=0; ii2<3; ii2++){
            for(ii3=0; ii3<3; ii3++){
                if(b[ii1] == a[ii2][ii3])
                    a[ii2][ii3] = -1;
            }
        }
    }

    if((a[0][0]==-1 && a[1][1]==-1 && a[2][2] == -1) || (a[0][2] ==-1 && a[1][1]==-1
        && a[2][0]==-1)){
            printf("Yes");
            return 0;
        }

    for(ii1=0; ii1<3; ii1++){
        for(ii2=0; ii2<3; ii2++){
            chk[ii1] += a[ii1][ii2];
        }
    }
    for(ii1=0; ii1<3; ii1++){
        for(ii2=0; ii2<3; ii2++){
            chk[3 + ii2] += a[ii1][ii2];
        }
    }

    for(ii1=0; ii1<9; ii1++){
        if(chk[ii1] == -3){
            printf("Yes");
            return 0;
        }
    }

    printf("No");
    return 0;
}