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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const double PI = 4*atan(1);
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-9;
const string YES = "Possible";
const string NO = "Impossible";
#define pb push_back
#define mp make_pair

int g[3][3];
int N;
int b[20];
bool f[3][3];

void solve()
{
	for(int i=0;i<N;++i){
		for(int j=0;j<3;++j)for(int k=0;k<3;++k){
			if(b[i]==g[j][k]) f[j][k]=true;
		}
	}
	for(int i=0;i<3;++i){
		bool fl = true;
		for(int j=0;j<3;++j){
			fl &= f[i][j];
		}
		if(fl){
			cout << "Yes\n";
			return;
		}
	}
	for(int j=0;j<3;++j){
		bool fl = true;
		for(int i=0;i<3;++i){
			fl &= f[i][j];
		}
		if(fl){
			cout << "Yes\n";
			return;
		}
	}
	if(f[0][0]&&f[1][1]&&f[2][2]) cout << "Yes\n";
	else if(f[0][2]&&f[1][1]&&f[2][0]) cout << "Yes\n";
	else cout << "No\n";
}

int main()
{
	ios::sync_with_stdio(false);
	for(int i=0;i<3;++i)for(int j=0;j<3;++j)cin >> g[i][j];
	cin >> N;
	for(int i=0;i<N;++i)cin >> b[i];
	solve();
	return 0;
}