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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define pi 3.14159
using namespace std;

int main()
{
	long long int n,i,a[3][3];
	int f=0,colsum=0,rowsum=0,k,j;
	for(i=0;i<3;i++)
		for(j=0;j<3;j++)
			cin>>a[i][j];
	cin>>n;
	int b[n];
	for(i=0;i<n;i++)
		cin>>b[i];
	for(k=0;k<n;k++)
		{
			for(i=0;i<3;i++)
				{
					rowsum=0;
					for(j=0;j<3;j++)
						{
							if(a[i][j]==b[k])
								a[i][j]=-1;
								rowsum+=a[i][j];
						}
					if(rowsum==-3)
					{
						f=1;
						break;
					}
				}
		}

for(k=0;k<n;k++)
		{
			for(j=0;j<3;j++)
				{
					colsum=0;
					for(i=0;i<3;i++)
						{
							if(a[i][j]==b[k])
								a[i][j]=-1;
								colsum+=a[i][j];
						}
					if(colsum==-3)
					{
						f=1;
						break;
					}
				}
		}
	if(a[0][0]+a[1][1]+a[2][2]==-3)
		f=1;
		
	if(a[0][2]+a[1][1]+a[2][0]==-3)
		f=1;
	if(f==1)
		cout<<"Yes"<<endl;
	else
		cout<<"No"<<endl;
		
		
}