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(void){
    int a[3][3];
    int f[3][3];
    int n;
    for(int i=0; i<3; i++){
        for(int j=0; j<3; j++){
            scanf("%d",&a[i][j]);
            f[i][j]=0;
        }
    }
    scanf("%d",&n);
    int b[n];
    for(int i=0; i<n; i++){
        scanf("%d",&b[i]);
        for(int j=0; j<3; j++){
            for(int k=0; k<3; k++){
                if(b[i]==a[j][k]){
                    f[j][k]=1;
                }
            }
        }
    }
    if(f[0][0]==1&&f[0][1]==1&&f[0][2]==1 ||
    f[0][0]==1&&f[1][0]==1&&f[2][0]==1 ||
    f[0][0]==1&&f[1][1]==1&&f[2][2]==1 ||
    f[0][1]==1&&f[1][1]==1&&f[2][1]==1 ||
    f[0][2]==1&&f[1][2]==1&&f[2][2]==1 ||
    f[1][0]==1&&f[1][1]==1&&f[1][2]==1 ||
    f[2][0]==1&&f[2][1]==1&&f[2][2]==1 ||
    f[2][0]==1&&f[1][1]==1&&f[0][2]==1){
        printf("Yes\n");
    } 
    else{
        printf("No\n");
    }
}