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 have_data(long data, long* p, long len);

int main(void){

    long a[4][4];
    for(long i=0;i<3;i++){
        for(long j=0;j<3;j++){
            scanf("%ld", &a[i][j]);
        }
    }
    for(long i=0;i<3;i++){
        a[i][3] = 1;
        a[3][i] = 1;
    }

    long n;
    scanf("%ld", &n);

    long b[n];
    for(long i=0;i<n;i++){
        scanf("%ld", &b[i]);
    }

    long d1 = 1, d2 = 1;
    for(long i=0;i<3;i++){
        for(long j=0;j<3;j++){
            a[i][3] &= have_data(a[i][j], b, n);
            a[3][j] &= have_data(a[i][j], b, n);
        }
        d1 &= have_data(a[i][i], b, n);
        d2 &= have_data(a[i][3-1-i], b, n);
    }

    long res = 0;
    for(long i=0;i<3;i++){
        res |= a[i][3] | a[3][i] | d1 | d2;
    }

    printf("%s\n", res ? "Yes" : "No");

    return 0;
}

int have_data(long data, long* p, long len){
    for(long i=0;i<len;i++){
        if(p[i] == data) return 1;
    }
    return 0;
}