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 n;
        int b[10];
        int i, j, k;

        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]);
        }

        for (i = 0; i < 3; i++) {
                for (j = 0; j < 3; j++) {
                        for (k = 0; k < n; k++) {
                                if (a[i][j] == b[k]) {
                                        a[i][j] = 0;
                                        break;
                                }
                        }
                }
        }

        if ( ((a[0][0] == 0) && (a[0][1] == 0) && (a[0][2] == 0)) ||
                        ((a[1][0] == 0) && (a[1][1] == 0) && (a[1][2] == 0)) ||
                        ((a[2][0] == 0) && (a[2][1] == 0) && (a[2][2] == 0)) ||
                        ((a[0][0] == 0) && (a[1][0] == 0) && (a[2][0] == 0)) ||
                        ((a[0][1] == 0) && (a[1][1] == 0) && (a[2][1] == 0)) ||
                        ((a[0][2] == 0) && (a[1][2] == 0) && (a[2][2] == 0)) ||
                        ((a[0][0] == 0) && (a[1][1] == 0) && (a[2][2] == 0)) ||
                        ((a[0][2] == 0) && (a[1][1] == 0) && (a[2][0] == 0)) )
        {
                printf("Yes\n");
        } else {
                printf("No\n");
        }

        return 0;
}