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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include<vector>
#include<algorithm>
#include<string>
using namespace std;
#define rep(N) for(int i=0;i<N;i++)
#define rep2(N1,N2) for(int i=0;i<N1;i++)for(int j=0;j<N2;j++)
#define rep3(N1,N2,N3)for(int i=0;i<N1;i++)for(int j=0;j<N2;j++)for(int k=0;k<N3;k++)
#define For(i,N) for(int i=0;i<N;i++)
void yn(bool b) {
	if (b)printf("yes\n");
	else printf("no\n");
}

void Yn(bool b) {
	if (b)printf("Yes\n");
	else printf("No\n");
}

void YN(bool b) {
	if (b)printf("YES\n");
	else printf("NO\n");
}

int main(){
	vector<vector<int>>A(3);
	rep(3)A[i].resize(3);
	rep2(3, 3)
		cin >> A[i][j];
	int N;
	cin >> N;
	vector<int>b(N);
	rep(N)cin >> b[i];
	For(i, N) {
		For(j, 3) {
			For(k, 3) {
				if (b[i] == A[j][k])
					A[j][k] = 0;
			}
		}
	}
	int flag = 0;
	int flag2 = 1;
	vector<int>vflag(8, 1);
	rep(3) {
		if (A[0][i] != 0)
			vflag[0] = 0;
		if (A[1][i] != 0)
			vflag[1] = 0;
		if (A[2][i] != 0)
			vflag[2] = 0;
		if (A[i][0] != 0)
			vflag[3] = 0;
		if (A[i][1] != 0)
			vflag[4] = 0;
		if (A[i][2] != 0)
			vflag[5] = 0;
		if (A[i][i] != 0)
			vflag[6] = 0;
		if (A[2 - i][i] != 0)
			vflag[7] = 0;
	}
	rep(8) {
		if (vflag[i] == 1)
			flag = 1;
	}
	Yn(flag);

}