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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int main(){

  int N,i,j,k,t;
  int A[3][3];
  int B[10];
  bool can;
  can=false;
  //  string S,A;
  
  for(i=0;i<3;i++){
    for(j=0;j<3;j++){
    cin>>A[i][j];
    }
  }
  cin>>N;
  for(i=0;i<N;i++){
    cin>>B[i];
  }

    for(i=0;i<3;i++){
    for(j=0;j<3;j++){
      // cout<<A[i][j]<<endl;
    }
  }
 
  for(i=0;i<3;i++){
    t=0;
    for(j=0;j<3;j++){
      for(k=0;k<N;k++){
	if(A[i][j]==B[k]){
	  t+=1;
	  if(t==3)can=true;
	  break;
	}
      }
    }
}
  
  for(i=0;i<3;i++){
    t=0;
    for(j=0;j<3;j++){
      for(k=0;k<N;k++){
	if(A[j][i]==B[k]){
	  t+=1;
	  if(t==3)can=true;
	  break;
	}
      }
    }
}

  t=0;
  for(i=0;i<3;i++){
      for(k=0;k<N;k++){
	if(A[i][2-i]==B[k]){
   t+=1;
   if(t==3)can=true;
      break;
	}
	if(t==3)can=true;
      }
   
}
  t=0;
for(i=0;i<3;i++){
  for(k=0;k<N;k++){
    if(A[i][i]==B[k]){
      t+=1;
      if(t==3)can=true;
      break;
    }
      }
    }

  if(can) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
  return 0;
}