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;
 
typedef long long ll;
typedef pair<int, int> pi;
 
 
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pi> vpi;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++);
#define F0R(i, a) for (int i=0; i<(a); i++);
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define all(x) x.begin(), x.end()
 
const int MOD = 1000000007;
const ll INF = 1e18;

 
int main() 
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	int M[10][10];
	for(int i=0; i<3; i++)
	{
		for(int j=0; j<3; j++)
		{
			cin>>M[i][j];
		}
	}
	cout<<"ok"<<endl;
	cout<<M[2][2]<<endl;
	
	int n; cin>>n;
	int N[11];
	for(int i=0; i<n; i++)
	{
		cin>>N[i];
	}
	for(int i=0; i<3; i++)
	{
		for(int j=0; j<3; j++)
		{
			for(int k=0;k<n;k++)
			{
				if(N[k]==M[i][j])
				{
					M[i][j]=-1;
				}
			}
		}
	}
	
	for(int i=0; i<3; i++)
	{
		if(M[i][1]==M[i][2] && M[i][2]==M[i][0] && M[i][0]==-1 )	
		{
			cout<<"Yes"<<endl;
			return 0;
		}
		if(M[1][i]==M[2][i] && M[0][i]==M[1][i] && M[0][i]==-1 )	
		{
			cout<<"Yes"<<endl;
			return 0;
		}
	}
	if(M[0][0]==-1 && M[1][1]==-1 && M[2][2]==-1 )
	{
		cout<<"Yes"<<endl;
		return 0;
	}
	if(M[0][2]==-1 && M[1][1]==-1 && M[2][0]==-1 )
	{
		cout<<"Yes"<<endl;
		return 0;
	}
	cout<<"No"<<endl;
	return 0;
}