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: Wrong Answer)

#include<iostream>
using namespace std;
int main()
{
    long long i,j,a[5][5],n,k,x,r[5]={0},c[5]={0},l,f=0,h=0;
    for(i=1;i<=3;i++)
    {
        for(j=1;j<=3;j++)
        {
            cin>>a[i][j];
        }
    }
    cin>>n;
    for(k=1;k<=n;k++)
    {
        cin>>x;
        for(i=1;i<=3;i++)
        {
            for(j=1;j<=3;j++)
            {
                if(a[i][j]==x)
                {
                    r[i]=r[i]+1;
                    c[j]=c[j]+1;
                    if(i==j)
                    {
                        h++;
                    }
                }
            }
        }
    }
    for(l=1;l<=3;l++)
    {
        if(r[l]==3||c[l]==3)
        {
            f++;
            cout<<"Yes"<<endl;
            break;
        }
    }
    if(f==0)
    {
        if(h==3)
        {
            cout<<"Yes"<<endl;
        }
        else
        {
            cout<<"No"<<endl;
        }
    }
    return 0;
}