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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t,r1=0,r2=0,r3=0,c1=0,c2=0,c3=0,d1=0,d2=0;
    int a[10];
    int n[11];
    for(int y=1;y<=9;y++)
    cin>>a[y];
    cin>>t;
    for(int x=1;x<=t;x++)
    {
    cin>>n[x];
      if(n[x]==a[1]||n[x]==a[2]||n[x]==a[3])
        r1++;
    else if(n[x]==a[4]||n[x]==a[5]||n[x]==a[6])
        r2++;
    else if(n[x]==a[7]||n[x]==a[8]||n[x]==a[9])
        r3++;
    if(n[x]==a[1]||n[x]==a[4]||n[x]==a[7])
        c1++;
    else if(n[x]==a[2]||n[x]==a[5]||n[x]==a[8])
        c2++;
    else if(n[x]==a[3]||n[x]==a[6]||n[x]==a[9])
        c3++;
    if(n[x]==a[1]||n[x]==a[5]||n[x]==a[9])
        d1++;
    else if(n[x]==a[3]||n[x]==a[5]||n[x]==a[7])
        d2++;
    }
  if(r1==3||r2==3||r3==3||c1==3||c2==3||c3==3||d1==3||d2==3)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
    return 0;
}