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()
{
 ios_base::sync_with_stdio(false);
 cin.tie(NULL);cout.tie(NULL);
 int a[4][4]={0},n,var=0,var1=0,temp=0;
 vector<int> q;
 for(int i=0;i<3;i++)
 {
     for(int j=0;j<3;j++)
     {
         cin>>a[i][j];
     }
 }
 cin>>n;
 for(int k=0;k<n;k++)
 {
     cin>>var;
     for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
                {
                if(var==a[i][j])
                a[i][j]=0;
                }
        }
 }
 for(int i=0;i<3 && temp==0;i++)
 {
     var=0;
     var1=0;
     for(int j=0;j<3;j++)
     {
         var+=a[i][j];
         var1+=a[j][i];
     }
     if(var==0 || var1==0)
     {
         temp=1;
     }
 }
 var=0;
 var1=0;
 for(int i=0;i<3 && temp==0;i++)
 {
     var+=a[i][i];
     var1+=a[i][3-i-1];
 }
 if(var==0 || var1==0)
 {
     temp=1;
 }
 if(temp)
 cout<<"Yes";
 else
 cout<<"No";
 return 0;
}