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, n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;
const int INF = 1e9;

int main() {
  ll a[3][3];
  bool ab[3][3];
  ll n;
  rep(i, 3) rep(j, 3) cin >> a[i][j];
  rep(i, 3) rep(j, 3) ab[i][j] = false;

  cin >> n;
  ll t;
  rep(i, n) {
    cin >> t;
    rep(i, 3) rep(j, 3) if (a[i][j] == t) ab[i][j] = true;
  }

  if ((ab[0][0] && ab[0][1] && ab[0][2]) ||
      (ab[1][0] && ab[1][1] && ab[1][2]) ||
      (ab[2][0] && ab[2][1] && ab[2][2]) ||
      (ab[0][0] && ab[1][0] && ab[2][0]) ||
      (ab[0][1] && ab[1][1] && ab[2][1]) ||
      (ab[0][2] && ab[1][2] && ab[2][2]) ||
      (ab[0][0] && ab[1][1] && ab[2][2]) || (ab[0][2] && ab[1][1] && ab[2][0]))
    cout << "Yes" << endl;
  else
    cout << "No" << endl;

  // rep(i, 3) {
  //   rep(j, 3) { cout << ab[i][j] ? 1 : 0; }
  //   cout << endl;
  // }
  return 0;
}