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 <stdlib.h>


int main(void) {

	int A[3][3],N,b[100];
	int tate,yoko,naname1,naname2;
    int i,j,k;

    for(i=0;i<3;i++)
	   {scanf("%d %d %d",&A[i][0],&A[i][1],&A[i][2]);}
    scanf("%d",&N);
    for(i=0;i<N;i++) {scanf("%d",&b[i]);}

    //tate & yoko
    for(i=0;i<3;i++) {
    	tate=yoko=0;
    	for(j=0;j<3;j++){
    		for(k=0;k<N;k++){
    		   if(A[i][j]==b[k]) tate++;
    		   if(A[j][i]==b[k]) yoko++;
    		}
    	}
    	if((tate==3) || (yoko==3))
    	  {printf("Yes");return(0);}
    }

    //naname
    for(i=0;i<3;i++){
       naname1=naname2=0;
       for(k=0;k<N;k++) {
    	   if(A[i][i]   ==b[k]) naname1++;
    	   if(A[2-i][i] ==b[k]) naname2++;
       }
       if(naname1==3 || naname2==3)
       {printf("yes");return(0);}
    }
    printf("No");return(0);
 }