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)

#define _USE_MATH_DEFINES
#pragma region include

#include <iostream>
#include <iomanip>
#include <stdio.h>

#include <sstream>
#include <algorithm>
#include <iterator>
#include <cmath>
#include <complex>

#include <string>
#include <cstring>
#include <vector>
#include <bitset>

#include <queue>
#include <set>
#include <map>
#include <stack>
#include <list>

#include <ctime>
////
//#include <random>//
#pragma endregion //#include
/////////

#pragma region typedef
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
#pragma endregion //typedef
////定数
const int INF = (int)1e9;
const LL MOD = (LL)1e9 + 7;
const LL LINF = (LL)4e18 + 20;
const LD PI = acos(-1.0);
const double EPS = 1e-9; 
/////////
using namespace::std;

void solve() {
	int A[3][3];
	for (int i = 0; i < 3; ++i) {
		for (int j = 0; j < 3; ++j) {
			cin >> A[i][j];
		}
	}
	int N;
	cin >> N;
	int b;
	for (int cnt = 0; cnt < N; ++cnt) {
		cin >> b;
		for (int i = 0; i < 3; ++i) {
			for (int j = 0; j < 3; ++j) {
				if (A[i][j] == b) {
					A[i][j] = 999;
				}
			}
		}
	}
	////////////////////////
	//
	bool ans = false;

	for (int i = 0; ans ==  false && i < 3; ++i) {
		ans = true;
		for (int j = 0; ans && j < 3; ++j) {
			if (A[i][j] != A[i][(j + 1) % 3] ) {
				ans = false;
				break;
			}
		}
		
	}
	if (ans) {
		cout << "Yes" << endl;
		return;
	}
	////////
	ans = false;
	for (int i = 0; ans == false && i < 3; ++i) {
		ans = true;
		for (int j = 0; ans && j < 3; ++j) {
			if (A[j][i] != A[(j+1)%3][i]) {
				ans = false;
				break;
			}
		}

	}
	if (ans) {
		cout << "Yes" << endl;
		return;
	}
	//
	{
		ans = true;
		if (A[0][0] != A[1][1] || A[1][1] != A[2][2] || A[2][2] != A[0][0]) {
			ans = false;
		}
		if (ans) {
			cout << "Yes" << endl;
			return;
		}
		if (A[0][2] != A[1][1] || A[1][1] != A[2][0] || A[2][0] != A[0][2]) {
			ans = false;
		}
		if (ans) {
			cout << "Yes" << endl;
			return;
		}
	}
	{
		cout << "No" << endl;
	}
}

#pragma region main
signed main(void) {
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	std::cout << std::fixed;//小数を10進数表示
	cout << setprecision(16);//小数点以下の桁数を指定//coutとcerrで別	

	solve();
}
#pragma endregion //main()