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>

#define rep(i, a, b) for (int i = a; i < (int)(b); ++i)
#define rrep(i, a, b) for (int i = b - 1; i >= (int)(a); --i)

using namespace std;

using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;

int main() {
    int A[3][3];
    rep(i, 0, 9) { cin >> A[i/3][i%3]; }

    int N;
    cin >> N;

    bool B[3][3]; rep(i, 0, 9) { B[i/3][i%3] = false; }

    rep(i, 0, N) {
        int b;
        cin >> b;
        rep(j, 0, 9) {
            if (A[j/3][j%3] == b) { B[j/3][j%3] = true; }
        }
    }

    // rep(i, 0, 3) {
    //     rep(j, 0, 3) {
    //         cout << B[i][j] << " ";
    //     }
    //     cout << endl;
    // }

    bool clear = false;

    rep(i, 0, 3) {
        if (B[i][0] && B[i][1] && B[i][2]) { clear = true; break; }
        if (B[0][i] && B[1][i] && B[2][i]) { clear = true; break; }
    }
    if (B[0][0] && B[1][1] && B[2][2]) { clear = true; }
    if (B[0][2] && B[1][1] && B[2][0]) { clear = true; }

    cout << (clear ? "Yes" : "No") << endl;
    return 0;
}