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>

int main(){
  int i,j,q,N,k,a[3][3],c=0,s[3],t[3],x=0,y=0;
  
  for(i=0;i<3;i++){
    for(j=0;j<3;j++){
      scanf("%d",&a[i][j]);//9x9ができた確認済み
    }
  }
  
 scanf("%d",&N);        //sumi
 // printf("%d",N);
  int b[N];
  for(i=0;i<N;i++){
    scanf("%d",&b[i]);
 
  }          //sumi
  // printf("%d\n",b[4]);
  for(k=0;k<N;k++){ 
   for(i=0;i<3;i++){
    for(j=0;j<3;j++){
      if(a[i][j]==b[k])
        a[i][j]=0;
    }
   }
  }//sumi
 
 /*  for(i=0;i<3;i++){
    for(j=0;j<3;j++){
  printf("%d\n",a[i][j]);  //観測用
    }
  }*/
  for(i=0;i<3;i++){
    t[i]=0;s[i]=0;
  }
  
  for(i=0;i<3;i++){
    for(j=0;j<3;j++){
      s[i]+=a[i][j];
    }
  }
    if(s[1]==0||s[0]==0||s[2]==0)
      c++;
   for(i=0;i<3;i++){
    for(j=0;j<3;j++){
      t[i]+=a[j][i];
    }
   }
      if(t[1]==0||t[0]==0||t[2]==0)
         c++;
   for(i=0;i<3;i++){
     x+=a[i][i];
     y+=a[i][2-i];
   }
     if(x==0||y==0)
    c++;

     if(c>0)
  printf("Yes");
     else
       printf("No");
}