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(){
    //card
    vector<int> card(9);
    for(int i=0; i<9; i++){
        cin >> card.at(i);
    }
    
    //n, numbers
    int n, num;
    cin >> n;;
    //check
    for(int i=0; i<n; i++){
        cin >> num;
        for(int j=0; j<9; j++){
            if(num == card.at(j)){
                card.at(j) = 0;
            }
            
        }
    }
    string ans = "No";
    for(int i=0; i<3; i++){
        //line
        if(card.at(3*i)==card.at(3*i+1)&&card.at(3*i+1)==card.at(3*i+2)){
            ans = "Yes";
            break;
        }
        //column
        if(card.at(i)==card.at(i+3)&&card.at(i+3)==card.at(i+6)){
            ans = "Yes";
            break;
        }
        //cross
        if(card.at(0)==card.at(4)&&card.at(4)==card.at(8)){
            ans = "Yes";
            break;
        }
        if(card.at(2)==card.at(4)&&card.at(4)==card.at(6)){
            ans = "Yes";
            break;
        }
    }
    /*for(int i=0; i<9; i++){
        cout << card.at(i) << endl;
    }
    */
    cout << ans << endl;
}