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>

int main() {
    int A[3][3];
    int i,j;
    for(i = 0; i < 3; i++) {
        for(j = 0; j < 3; j++) {
            scanf("%d", &A[i][j]);
        }
    }  
 
    int N;
    scanf("%d", &N);

    int B[N];
    int k;
    for(k = 0; k < N; k++) {
        scanf("%d", &B[k]);

        for(i = 0; i < 3; i++) {
            for(j = 0; j < 3; j++) {
                if(B[k] == A[i][j]) {
                    A[i][j] = 0;
                }
            }
        } 

    } 

    int ans = 0;

    if((A[0][0] + A[0][1] + A[0][2]) == 0) ans = 1;
    else if((A[1][0] + A[1][1] + A[1][2]) == 0) ans = 1;
    else if((A[2][0] + A[2][1] + A[2][2]) == 0) ans = 1;
    else if((A[0][0] + A[1][0] + A[2][0]) == 0) ans = 1;
    else if((A[0][1] + A[1][1] + A[2][1]) == 0) ans = 1;
    else if((A[0][2] + A[1][2] + A[2][2]) == 0) ans = 1;
    else if((A[0][0] + A[1][1] + A[2][2]) == 0) ans = 1;
    else if((A[0][2] + A[1][1] + A[2][0]) == 0) ans = 1;
    else ans = 0;

    if(ans == 1) {
        printf("Yes");
    } else {
        printf("No");
    }

return 0;
}