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>
using namespace std;

std::vector<int> A(9);
int n;
string Message = "No";

int main() {
  for (size_t i = 0; i < 9; i++) {
    cin >> A.at(i);
  }
  cin >> n;
  std::vector<int> B(n);
  for (size_t i = 0; i < n; i++) {
    cin >> B.at(i);
  }
  for (size_t i = 0; i < 9; i++) {
    for (size_t j = 0; j < n; j++) {
      if (A.at(i) == B.at(j)) {
        A.at(i) = 0;
      }
    }
  }
  for (size_t i = 0; i < 9; i++) {
    if (A.at(i) != 0) {
      A.at(i) = 1;
    }
  }
  for (size_t i = 0; i < 3; i++) {
    if (A.at(i) + A.at(i + 1) + A.at(i + 2) == 0) {
      Message = "Yes";
    }
  }
  for (size_t i = 0; i < 3; i++) {
    if (A.at(i) + A.at(i + 3) + A.at(i + 6) == 0) {
      Message = "Yes";
    }
  }
  if (A.at(0) + A.at(4) + A.at(8) == 0) {
    Message = "Yes";
  }
  if (A.at(2) + A.at(4) + A.at(6) == 0) {
      Message = "Yes";
  }
  cout << Message << endl;
  return 0;
}