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<cstring>
using namespace std;
int main(){
	int bingo[5][5],v[5][5],n,b[15];
	memset(v,0,sizeof(v));
	for(int i=1;i<=3;i++)
		for(int j=1;j<=3;j++)
			cin>>bingo[i][j];
	cin>>n;
	for(int i=0;i<n;i++) cin>>b[i];
	for(int i=0;i<n;i++){
		for(int j=1;j<=3;j++){
			for(int k=1;k<=3;k++){
				if(bingo[j][k]==b[i]) v[j][k]=1;
			}
		}
	}
	bool f=false;
	int m,x;
	for(int i=1;i<=3;i++){
		m=0;x=0;
		for(int j=1;j<=3;j++){
			if(v[i][j]==1) m++;
			if(m==3){
				f=true;
				break;
			}
			if(v[j][i]==1) x++;
			if(x==3){
				f=true;
				break;
			}
		}
	}
	if(v[1][1]==1&&v[2][2]==1&&v[3][3]==1||v[1][3]==1&&v[2][2]==1&&v[3][1]==1) f=true;
	if(f) cout<<"Yes";
	else cout<<"No";
	return 0;
}