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 <string>
#include <vector>
#include <algorithm>
#include <map>
#include <math.h>

#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;

int main() {
	vector<vector<int>>A(3,vector<int>(3));
	vector<vector<int>>B(3,vector<int>(3,0));
	int n;
	rep(i,3){
		rep(j,3){
			cin >> A[i][j];
		}
	}
	cin >> n;
	vector<int>C(n);
	int x;
	rep(i,n){
		cin >> C[i];
	}

	rep(i,n){
		rep(j,3){
			rep(k,3){
				if(C[i] == A[j][k]){
					B[j][k] = 1;
				}
			}
		}
	}
	string ans = "No";
	rep(i,3){
		if(B[i][0] == 1 && B[i][1] == 1 && B[i][2] == 1){
			ans = "Yes"; 
			break;
		}
		if(B[0][i] == 1 && B[1][i] == 1 && B[2][i] == 1){
			ans = "Yes"; 
			break;
		}
	}
	if(B[0][0] == 1 && B[1][1] == 1 && B[2][2] == 1) ans = "Yes";
	if(B[2][0] == 1 && B[1][1] == 1 && B[0][2] == 1) ans = "Yes";
	cout << ans;
	return 0;
}