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 9.2.1) (judge verdict: Accepted)

#define _GLIBCXX_DEBUG
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ALL(v) v.begin(), v.end()
typedef long long ll;

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

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  vector<int> A(9),B(9,0);
  rep(i,9) cin>>A[i];

  int n;
  cin>>n;
  
  int x;
  rep(i,n){
    cin>>x;
    rep(j,9){
      if(A[j]==x) B[j]=1;
    }
  }
  
  bool b=false;
  if(B[0]==1 && B[1]==1 && B[2]==1) b=true;
  if(B[3]==1 && B[4]==1 && B[5]==1) b=true;
  if(B[6]==1 && B[7]==1 && B[8]==1) b=true;
  if(B[0]==1 && B[3]==1 && B[6]==1) b=true;
  if(B[1]==1 && B[4]==1 && B[7]==1) b=true;
  if(B[2]==1 && B[5]==1 && B[8]==1) b=true;
  if(B[0]==1 && B[4]==1 && B[8]==1) b=true;
  if(B[2]==1 && B[4]==1 && B[6]==1) b=true;
  
  if(b) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
  
  return 0;
}