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 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int main()
{
	int a[3][3][2] = {0};
	for(int i = 0; i < 3; i++)
	{
		for(int j = 0; j < 3; j++)
		{
			scanf("%d", &a[i][j][0]);
		}
	}
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i++)
	{
		int x, found = 0; scanf("%d", &x);
		for(int j = 0; j < 3; j++)
		{
			for(int k = 0; k < 3; k++)
			{
				if(x == a[j][k][0])
				{
					a[j][k][1] = 1;
					found = 1;
					break;
				}
			}
			if(found) break;
		}
	}
	int yes = 0, rcount[3] = {0}, ccount[3] = {0}, d1 = 0, d2 = 0;
	for(int i = 0; i < 3; i++)
	{
		for(int j = 0; j < 3; j++)
		{
			if(a[i][j][1] == 1) 
			{
				rcount[i]++;
				ccount[j]++;
			}
			if(i == j && a[i][j][1] == 1) d1++;
			else if (i + j == 2 && a[i][j][1] == 1) d2++;
		}
	}
	for(int i = 0; i < 3; i++)
	{
		if(rcount[i] == 3 || ccount[i] == 3)
		{
			yes = 1;
			break;
		}
	}
//	printf("%d %d %d\n", yes, d1, d2);
	(yes || d1 == 3 || d2 == 3)? puts("Yes\n") : puts("No\n");
	return 0;
}