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: Wrong Answer)

#include <bits/stdc++.h>
#include <math.h>
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;
using ll = long long;

int main() {
  vector<vector<int>> a(3, vector<int>(3));
  rep(i, 3) {
    rep(j, 3) cin >> a.at(i).at(j);
  }
  int n;
  cin >> n;
  vector<int> b(n);
  rep(i, n) cin >> b.at(i);
  vector<vector<bool>> appear(3, vector<bool>(3));
  rep(i, 3) {
    rep(j, 3) appear.at(i).at(j) = false;
  }
  rep(i, 3) {
    rep(j, 3) {
      rep(k, n) {
        if(appear.at(i).at(j) == b.at(k)) appear.at(i).at(j) = true;
      }
    }
  }
  rep(i, 3) {
    if (appear.at(i).at(0) == true && appear.at(i).at(1) == true && appear.at(i).at(2) == true) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  rep(i, 3) {
    if (appear.at(0).at(i) == true && appear.at(1).at(i) == true && appear.at(2).at(i) == true) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  if (appear.at(0).at(0) == true && appear.at(1).at(1) == true && appear.at(2).at(2) == true) {
      cout << "Yes" << endl;
      return 0;
  }
  else if (appear.at(0).at(2) == true && appear.at(1).at(1) == true && appear.at(2).at(0) == true) {
      cout << "Yes" << endl;
      return 0;
  }
  else cout << "No" << endl;
}