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 rep(i,N) for (int i = 0; i < (N); ++i)
#define all(v) v.begin(), v.end()
using namespace std;
using ll = long long;
using P = pair<int, int>;
int main(void){
  int A[3][3];
  int N;
  cin >> N;
  int b[N];
  rep(i,3){
    rep(j,3){
      cin >> A[i][j];
    }
  }
  rep(i,N) cin >> b[i];

  bool appear[3][3];
  rep(i,3){
    rep(j,3){
      appear[i][j] = false;
      rep(k,N){
        if (A[i][j] == b[k]) appear[i][j] = true;
      }
    }
  }

  string ans = "No";
  rep(i,3){
    if (appear[i][0] && appear[i][1] && appear[i][2]) ans = "Yes";
    if (appear[0][i] && appear[1][i] && appear[2][i]) ans = "Yes";
  }
  if (appear[0][0] && appear[1][1] && appear[2][2]) ans = "Yes";
  if (appear[0][2] && appear[1][1] && appear[2][0]) ans = "Yes";

  cout << ans << endl;
}