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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
	vector<uint16_t> A;
	for (size_t i = 0; i < 9; ++i)
	{
		uint16_t a_ij;
		cin >> a_ij;
		A.push_back(a_ij);
	}

	size_t N;
	cin >> N;
	if (N >= 3)
	{
		vector<uint16_t>::iterator itr;
		uint16_t hit(0);
		for (size_t i = 0; i < N; ++i)
		{
			uint16_t bn(0);
			cin >> bn;
			itr = find(A.begin(), A.end(), bn);
			if (itr != A.end())
				hit |= 1 << distance(A.begin(), itr);
		}
		const uint16_t tate(1 | 1 << 3 | 1 << 6);
		const uint16_t yoko(1 | 1 << 1 | 1 << 2);
		const uint16_t naname1(1 | 1 << 4 | 1 << 8);
		const uint16_t naname2(4 | 1 << 4 | 1 << 6);

		for (size_t i = 0; i < 3; ++i)
		{
			if ((hit & (tate << i))
				|| (hit & (yoko << i)))
			{
				cout << "Yes" << endl;
				return 0;
			}
		}
		if ((hit & naname1)
			|| (hit & naname2))
		{
			cout << "Yes" << endl;
			return 0;
		}
	}
	cout << "No" << endl;

	return 0;
}