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 <bits/stdc++.h>
using namespace std;

int main() {
  int N,A,B,C;
  int X[3][3],Y[3][3];
  Y[0][0]=0;
    Y[0][1]=0;
    Y[0][2]=0;
    Y[1][0]=0;
    Y[1][1]=0;
    Y[1][2]=0;
    Y[2][0]=0;
    Y[2][1]=0;
    Y[2][2]=0;
  cin>>X[0][0]>>X[0][1]>>X[0][2]>>X[1][0]>>X[1][1]>>X[1][2]>>X[2][0]>>X[2][1]>>X[2][2];
  cin >>N;
  for(int i=0;i<N;i++){
    cin>>B;
    for(int j=0;j<3;j++){
      for(int u=0;u<3;u++){
        if(X[j][u]==B){
          Y[j][u]=1;
        }
      }
    }
  }
  if((Y[0][0]==1&&Y[0][1]==1&&Y[0][2]==1)||(Y[1][0]==1&&Y[1][1]==1&&Y[1][2]==1)||(Y[2][0]==1&&Y[2][1]==1&&Y[2][2]==1)||(Y[0][0]==1&&Y[1][1]==1&&Y[2][2]==1)||(Y[2][0]==1&&Y[1][1]==1&&Y[0][2]==1)||(Y[2][0]==1&&Y[1][0]==1&&Y[0][0]==1)||(Y[2][1]==1&&Y[1][1]==1&&Y[0][1]==1)||(Y[0][2]==1&&Y[1][2]==1&&Y[2][2]==1)){
  cout <<"Yes"<< endl;
}
  else{
    cout<<"No"<<endl;
  }
}