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 <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <algorithm>

int N, A[3][3], b[10];
bool appear[3][3];

using namespace std;

int main()
{
	for (int i = 0; i < 3; i++) {
		for (int j = 0; j < 3; j++) {
			cin >> A[i][j];
		}
	}
	cin >> N;

	for (int i = 0; i < 3; i++) {
		cin >> b[i];
	}

	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			appear[i][j] = false;
			for (int k = 0; k < N; k++) {
				if (A[i][j] == b[k]) {
					appear[i][j] = true;
				}
			}
		}
	}
	string ans = "No";
	for (int i = 0; i < 3; i++) {
		if (appear[i][0] && appear[i][1] && appear[i][2]) {
			ans = "Yes";
		}
	}
	for (int i = 0; i < 3; i++) {
		if (appear[0][i] && appear[1][i] && appear[2][i]) {
			ans = "Yes";
		}
	}
	if (appear[0][0] && appear[1][1] && appear[2][2]) {
		ans = "Yes";
	}
	if (appear[0][2] && appear[1][1] && appear[2][0]) {
		ans = "Yes";
	}
	cout << ans << endl;
}