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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
int a[3][3], n , b[10];
bool appear[3][3];

int main(){
    int count = 0;
    for(int i = 0; i < 3; i++){
        for(int j = 0; j < 3; i++){
            cin >> a[i][j];
        }
    }
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> b[i];
    }
    for(int i = 0; i < 3; i++){
        for(int j = 0; j < 3; j++){
            appear[i][j] = false;
            for(int k=0;k < n;k++){
                if(b[k]==a[i][j])   appear[i][j] = true;
        }
     }
    }
    for(int i = 0;i < 3;i++){
        if(appear[i][0]==appear[i][1]&&appear[i][0]==appear[i][2]){
             count = 1;
             break;
        }
    }
    for(int i = 0;i < 3;i++){
        if(appear[0][i]==appear[1][i]&&appear[0][i]==appear[2][i]){
        count = 1;
        break;
        }
    }
    if(appear[0][2]==appear[1][1]&&appear[1][1]==appear[2][0]){
     count = 1;
    }
    if(count == 0){
        cout << "No" <<endl;
    }else{
     cout << "Yes" << endl;
    }
    return 0;
}