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[9], n, b[10], i, k, an[9];
	
	for(i = 0; i < 9; i++){
		an[i] = 0;
	}
	
	for(i = 0; i < 9; i++){
		scanf("%d", &a[i]);
	}
	scanf("%d", &n);
	for(i = 0; i < n; i++){
		scanf("%d", &b[i]);
	}
	
	for(i = 0; i < n; i++){
		for(k = 0; k < 9; k++){
			if(b[i] == a[k]){
				an[k] = 1;
			}
		}
	}
	/*
	i = 0;
	while(i < 9){
		printf("%d %d %d\n", an[i], an[i + 1], an[i + 2]);
		i += 3;
	}
	*/
	if((an[0] == 1 && an[1] == 1 && an[2] == 1) ||
	   (an[3] == 1 && an[4] == 1 && an[5] == 1) ||
	   (an[6] == 1 && an[7] == 1 && an[8] == 1) ||
	   (an[0] == 1 && an[3] == 1 && an[6] == 1) ||
	   (an[1] == 1 && an[4] == 1 && an[7] == 1) ||
	   (an[2] == 1 && an[5] == 1 && an[8] == 1) ||
	   (an[0] == 1 && an[4] == 1 && an[8] == 1) ||
	   (an[2] == 1 && an[4] == 1 && an[6] == 1)){
		printf("Yes");
	}
	else{
		printf("No");
	}
	
	return 0;
}