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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <cmath>
using namespace std;


int a[3][3];

void solve()
{
	for (int i = 0; i < 3; i++)
	{
		for (int k = 0; k < 3; k++)
		{
			cin >> a[i][k];
		}
	}


	int m;
	cin >> m;
	for (int i = 0; i < m; i++)
	{
		int x;
		cin >> x;

		for (int i = 0; i < 3; i++)
		{
			for (int k = 0; k < 3; k++)
			{
				if (a[i][k] == x)
				{
					a[i][k] = -1;
				}
			}
		}
	}

	bool yes = false;

	for (int i = 0; i < 3; i++)
	{
		bool t1 = true, t2 = true;
		for (int k = 0; k < 3; k++)
		{
			if (a[i][k] != -1)
			{
				t1 = false;
			}

			if (a[k][i] != -1)
			{
				t2 = false;
			}

			yes = max(yes, max(t1, t2));
		}


		bool d1 = true, d2 = true;
		for (int i = 0; i < 3; i++)
		{
			if (a[i][i] != -1)
			{
				d1 = false;
			}

			if (a[i][3 - i - 1] != -1)
			{
				d2 = false;
			}
		}

		yes = max(yes, max(d1, d2));
	}

	if (yes)
	{
		cout << "Yes";
	}
	else
	{
		cout << "No";
	}
}

int main()
{
	solve();
}