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;
#define pi 3.14159265359
#define int long long
#define pii pair<int,int>
const int mod=1e9+7;

int dx[]={-1,0,1,0};
int dy[]={0,-1,0,1};




int32_t  main()
{ 
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);




int a[5][5]={0};
map<int,pii>mp;
for(int i=1;i<=3;i++)for(int j=1;j<=3;j++){ cin>>a[i][j];mp[a[i][j]]={i,j};}
int n;
cin>>n;

for(int i=1;i<=n;i++)
{
  int x;
  cin>>x;
  if(mp.find(x)!=mp.end())
  {
    a[mp[x].first][mp[x].second]=-1;
  }
}

for(int i=1;i<=3;i++)
{
  for(int j=1;j<=3;j++)
  {
     if(a[i][j]==-1)
     {
       if(a[i-1][j]==-1&&a[i+1][j]==-1)
       {
         cout<<"Yes\n"; return 0;
       }
       if(a[i][j-1]==-1&&a[i][j+1]==-1)
       {
         cout<<"Yes\n"; return 0;
       }
       if(a[i-1][j-1]==-1&&a[i+1][j+1]==-1)
       {
         cout<<"Yes\n"; return 0;
       }
       if(a[i-1][j+1]==-1&&a[i+1][j-1]==-1)
       {
          cout<<"Yes\n"; return 0;
       }
     }
  }
}
cout<<"No\n";



  
}