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 <iostream>
#include <vector>
#include <array>
#include <map>

using namespace std;

int main()
{
    map<int, int> a;

    // オリジナルの回答
    for (int i = 0; i < 3; i++) {
        int input;
        
        cin >> input;
        a[input] = i * 3;

        cin >> input;
        a[input] = i * 3 + 1;

        cin >> input;
        a[input] = i * 3 + 2;
    }

    int n;
    cin >> n;

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

    vector<bool> temp(9);
    fill(temp.begin(), temp.end(), false);

    for (int i = 0; i < b.size(); i++) {
        if (a.count(b[i]) != 0) {
            temp[a.at(b[i])] = true;
        }
    }

    bool isBingo = false;
    for (int i = 0; i < 3; i++) {
        isBingo |= temp.at(i * 3) && temp.at(i * 3 + 1) && temp.at(i * 3 + 2);
        isBingo |= temp[i] && temp.at(i + 3) && temp.at(i + 6);
    }
    
    isBingo |= temp[0] && temp[4] && temp[8];
    isBingo |= temp[2] && temp[4] && temp[6];

    cout << (isBingo ? "Yes" : "No") << endl;

    return 0;
}