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 a11, a12, a13, a21, a22, a23, a31, a32, a33;
    int n, i, j;
    int p[100];

    scanf("%d%d%d", &a11, &a12, &a13);
    scanf("%d%d%d", &a21, &a22, &a23);
    scanf("%d%d%d", &a31, &a32, &a33);

    scanf("%d", &n);

    for (i = 0; i < n; i++)
    {
        scanf("%d", &p[i]);
    }

    for (j = 0; j < n; j++)
    {
        if(a11 == p[j]){
            a11 = 0;
        }

        if(a12 == p[j]){
            a12 = 0;
        }

        if(a13 == p[j]){
            a13 = 0;
        }

        if(a21 == p[j]){
            a21 = 0;
        }

        if(a22 == p[j]){
            a22 = 0;
        }

        if(a23 == p[j]){
            a23 = 0;
        }

        if(a31 == p[j]){
            a31 = 0;
        }

        if(a32 == p[j]){
            a32 = 0;
        }

        if(a33 == p[j]){
            a33 = 0;
        }
    }

    if((a11==a12&&a12==a13&&a13==0) || (a21==a22&&a22==a23&&a23==0) || (a31==a32&&a32==a33&&a33==0) || (a11==a21&&a21==a31&&a31==0) || (a12==a22&&a22==a32&&a32==0) || (a13==a23&&a23==a33&&a33==0) || (a11==a22&&a22==a33&&a33==0) || (a13==a22&&a22==a31&&a31==0)){
        printf("Yes");
    }

    else{
        printf("No");
    }
}