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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define INF 1000000007
#define ll long long
#define F first
#define S second
#define pii pair<ll int,ll int>
#define pb push_back
#define ppb pop_back
#define vi vector<ll int>



int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t=1;
	//cin >> t;
	while(t--){
		int A[3][3];
		for(int i=0;i<3;i++)
			for(int j=0;j<3;j++) cin >> A[i][j];
		int n;
		cin >> n;
		int b;
		int f[3][3]={0};
		for(int k=0;k<n;k++){
			cin >> b;
			for(int i=0;i<3;i++){
				for(int j=0;j<3;j++){
					if(b == A[i][j]) f[i][j]=1;
				}
			}
		}
		bool flag=false;
		for(int i=0;i<3;i++){
			int r=0;
			for(int j=0;j<3;j++){
				r += A[i][j];
			}
			if(r==3){
				flag=true;
				break;
			}
		}
		for(int i=0;i<3;i++){
			int r=0;
			for(int j=0;j<3;j++){
				r += A[j][i];
			}
			if(r==3){
				flag=true;
				break;
			}
		}
		if(f[0][0]==1 && f[1][1]==1 && f[2][2]==1) flag=true;
		if(f[0][2]==1 && f[1][1]==1 && f[2][0]==1) flag=true;
		if(flag) cout << "YES";
		else cout << "NO";
	}
	return 0;
}