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>

using namespace std;

int main()
{
    int b[3][3];
    //int a[3][3];
    int N = 0;
    int num[10];

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

    cin >> N;
    
    if (N < 3){
        cout << "No" << endl;
        return 0;
    }

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

    for (int i = 0; i < 3; i++){
        for (int j = 0; j < 3; j++){
            for (int k = 0; k < N; k++){
                if (b[i][j] == num[k])
                {
                    b[i][j] = -1;
                    //cout << b[i][j] << endl;
                }
            }
        }
    }

    for(int i = 0 ; i < 3; i++){
        int c = 1;
        for (int j = 0 ; j < 3 ; j++){
            c *= b[i][j];
        }
        if (c == -1){
            cout << "Yes" << endl;
            return 0;
        }
    }

    for(int i = 0 ; i < 3; i++){
        int c = 1;
        for (int j = 0 ; j < 3 ; j++){
            c *= b[j][i];
        }
        if (c == -1){
            cout << "Yes" << endl;
            return 0;
        }
    }

    if (b[0][0] * b[1][1] * b[2][2] == -1){
        cout << "Yes" << endl;
        return 0;
    }

    cout << b[0][0] << endl;

    cout << "No" << endl;
    return 0;
}

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