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: Accepted)

#include <stdio.h>
#include <math.h>
int main(void)
{
  int A[3][3];
  
  for(int i = 0; i < 3; i++){
    for(int j = 0; j < 3; j++){
      scanf("%d", &A[i][j]);
    }
  }
    
  int N = 0;
   
  scanf("%d", &N);
 
  int b[N];
  
  for(int i = 0; i < N; i++){
    scanf("%d", &b[i]);
  }
  
  int ans1[3] = {0, 0, 0};
  int ans2[3] = {0, 0, 0};
  
  for(int i = 0; i < 3; i++){
    for(int j = 0; j < 3; j++){
      for(int k = 0; k < N; k ++){
        if(A[i][j] == b[k]){
          ans1[i] ++;
        }
      }
    }
  }
  
  for(int i = 0; i < 3; i++){
    for(int j = 0; j < 3; j++){
      for(int k = 0; k < N; k ++){
        if(A[j][i] == b[k]){
          ans2[i] ++;
        }
      }
    }
  }
  
  int ans3 = 0;
  
  for(int i = 0; i < 3; i++){
    for(int j = 0; j < N; j++){
      if(A[i][i] == b[j]){
          ans3 ++;
      }
    }
  }
  
  int ans4= 0;
    
  for(int i = 0; i < 3; i++){
    for(int j = 0; j < N; j++){
      if(A[i][2-i] == b[j]){
          ans4 ++;
      }
    }
  }  
   
  if(ans1[0] == 3 || ans1[1] == 3 ||ans1[2] == 3 ||ans2[0] == 3 ||ans2[1] == 3 ||ans2[2] == 3 ||ans3 == 3 ||ans4 == 3){
    printf("Yes");
  }else{
    printf("No");
  }
   
   
  return 0;
}