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<iostream>
using namespace std;

int main(){
  int A [3][3];
  for(int i=0;i<3;i++){
    for(int r=0;r<3;r++){
      cin >> A[i][r];
    }
  }
  int N;
  cin >> N;
  int B[N];
  for(int i=0;i<N;i++)cin >> B[i];
  for(int i=0;i<2;i++){
    for(int r=0;r<2;r++){
      for(int t=0;t<N;t++){
        if(A[i][r] == B[t])A[i][r]=0;
      }
    }
  }
  int a[3]={0,0,0};
  int b[3]={0,0,0};
  int c,d;
 
  c = A[0][0]+A[1][1]+A[2][2];
  d = A[0][2]+A[1][1]+A[2][0];
  for(int i=0;i<3;i++){
    for(int r=0;r<3;r++){
      a[i] += A[i][r];
      b[i] += A[r][i];
    }
  }
  int s=0;
  for(int i=0;i<3;i++){
    if(a[i]==0)s=1;
    if(b[i]==0)s=1;
  }
  if(c==0 || d==0)s==1;
  if(s==1)cout << "Yes" << endl;
  if(s==0)cout << "No" << endl;
}