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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
  int num[9];
  int atk[9]={0,0,0,0,0,0,0,0,0};
  int N;
  scanf("%d %d %d", &num[0], &num[1], &num[2]);
  scanf("%d %d %d", &num[3], &num[4], &num[5]);
  scanf("%d %d %d", &num[6], &num[7], &num[8]);
  scanf("%d", &N);
  
  int re;
  for(int i=0; i<N; i++){
    //printf("%d %d %d\n", atk[0], atk[3], atk[6]);
    scanf("%d", &re);
    for(int j=0; j<9; j++){
      if(num[j]==re){
        atk[j]=1;
      }
    }
    if(atk[0]==1){
      if(atk[3]*atk[6]==1){
        printf("YES");
        return 0;
      }else if(atk[4]*atk[8]==1){
        printf("YES");
        return 0;
      }else if(atk[1]*atk[2]==1){
        printf("YES");
        return 0;
      }
    }
    if(atk[2]==1){
      if(atk[5]*atk[8]==1){
        printf("YES");
        return 0;
      }else if(atk[4]*atk[6]==1){
        printf("YES");
        return 0;
      }
    }
    if(atk[6]*atk[7]*atk[8]==1){
      printf("YES");
      return 0;
    }
    if(atk[1]*atk[7]*atk[4]==1){
      printf("YES");
      return 0;
    }
    if(atk[3]*atk[4]*atk[5]==1){
      printf("YES");
      return 0;
    }
  }
  printf("NO");
  
  return 0;
}