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;
int a[3][3];
int n;

int main() {
    for(int i = 0; i < 3; ++i) {
        for(int j = 0; j < 3; ++j) {
            cin >> a[i][j];
        }
    }
    vector<vector<bool>> abo(3, vector<bool>(3, false));
    cin >> n;
    for(int i = 0; i < n; ++i) {
        int b;
        cin >> b;
        for(int j = 0; j < 3; ++j) {
            for(int k = 0; k < 3; ++k) {
                if(b == a[j][k]) {
                    abo.at(j).at(k) = true;
                }
            }
        }
    }

    bool flg = false;

    for(int i = 0; i < 3; ++i) {
        if(abo.at(i).at(0) == true && abo.at(i).at(1) == true && abo.at(i).at(2) == true) {
            flg = true;
        }
    }

    for(int i = 0; i < 3; ++i) {
        if(abo.at(0).at(i) == true && abo.at(1).at(i) == true && abo.at(2).at(i) == true) {
            flg = true;
        }
    }

    if(abo.at(0).at(0) == true && abo.at(1).at(1) == true && abo.at(2).at(2) == true) {
        flg = true;
    }

    if(abo.at(0).at(2) == true && abo.at(1).at(1) == true && abo.at(2).at(0) == true) {
        flg = true;
    }

    if(flg) {
        cout << "Yes" << endl;
    }else {
        cout << "No" << endl;
    }
}