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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define debug(x) cerr << #x << ": " << x << endl;

using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;

const int MAX { 1000010 };
const ll MOD { 1000000007 };
const double PI = acos(-1.0);

int main(){
	ios::sync_with_stdio(false);
    map<int, ii> c;
    for(int i=0; i<3; ++i)
        for(int j=0; j<3; ++j){
            int aux;
            cin >> aux;
            c[aux] = make_pair(i, j);
        }

    int N;
    int I[3] = {}, J[3] = {};
    cin >> N;
    while(N--){
        int aux;
        cin >> aux;
        if (c.find(aux) != c.end()){
            debug(aux);
            ii v = c[aux];
            debug(v.first);
            debug(v.second);
            I[v.first]++;
            J[v.second]++;
        }
    }

    int aux_dig = 0;
    for(int i=0; i<3; ++i){
        debug(I[i]);
        debug(J[i]);
        if(I[i] && J[i]) aux_dig++;

        if(I[i] == 3 || J[i] == 3){
            cout << "Yes\n";
            return 0;
        }
    }

    aux_dig == 3? cout << "Yes\n" : cout << "No\n";

    return 0;
}