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;
typedef long long ll;
ll mod = 1e9+7;
ll inf = 1e18;
int a[3][3];
bool aa[3][3];
int main() {
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            cin >> a[i][j];
        }
    }
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        int c;
        cin >> c;
        for (int j = 0; j < 3; j++) {
            for (int k = 0; k < 3; k++) {
                if (c == a[j][k]) aa[j][k] = 1;
            }
        }
    }

    bool f1 = 1, f2 = 1, f = 0;
    for (int i = 0; i < 3; i++) {
        f |= (aa[i][0] && aa[i][1] && aa[i][2]);
        f |= (aa[0][i] && aa[1][i] && aa[2][i]);
        f1 &= aa[i][i];
        f2 &= aa[i][2 - i];
    }
    // for (int i = 0; i < 3; i++) {
    //     for (int j = 0; j < 3; j++) {
    //         cout << aa[i][j];
    //     }
    //     cout << endl;
    // }
    cout << (f||f1||f2 ? "Yes":"No") << endl;
    return 0;
}