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>
#include <string.h>
int main(void)
{
        int a[3][3],b[20],m,n,i,j=0,k=0,l=0,p=0;
        char s[1000];

        for(i=0; i<3; i++){
                for(j=0; j<3; j++)
                scanf("%d",&a[i][j]);
        }
        scanf("%d",&n);

        for(i=0; i<n; i++)      scanf("%d",&b[i]);
                
        for(i=0; i<3; i++){ p=0; //ヨコline check
                for(j=0; j<3; j++){ 
                        for(k=0; k<n; k++)
                                if(a[i][j]==b[k]) p++;
                        if(p==3){ printf("Yes\n"); return 0;}
                }
        }
        
        for(i=0; i<3; i++){ p=0; //タテline check
                for(j=0; j<3; j++){ 
                        for(k=0; k<n; k++)
                                if(a[j][i]==b[k]) p++;
                        if(p==3){ printf("Yes\n"); return 0;}
                }
        }

                while(a[1][1]==b[i]){ i++;
                        if(i==111){ printf("No\n"); return 0;}
                }
p=0;
        for(i=0; i<3; i++){
                for(k=0; k<n; k++)
                if(a[i][i]==b[k]) p++;
                if(p==3){ printf("Yes\n"); return 0;}
        }
        p=0;
        for(i=0; i<3; i++){ 
                for(k=0; k<n; k++)
                if(a[2-i][i]==b[k])p++;
                if(p==3){ printf("Yes4\n");return 0;}
        }

          printf("No\n");
          return 0;
}