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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int[][] as = new int[3][3];
        for( int i = 0 ; i < 3 ; i++ ){
            String[] params = sc.nextLine().split(" ");
            for( int j = 0 ; j < 3 ; j++ ){
                as[i][j] = Integer.parseInt( params[j] );
            }
        }
        int n = Integer.parseInt( sc.nextLine() );
        int[] bs = new int[n];
        for( int i = 0 ; i < n ; i++ ){
            bs[i] = Integer.parseInt( sc.nextLine() );
            for( int j = 0 ; j < 3 ; j++ ){
                for( int k = 0 ; k < 3 ; k++ ){
                    if( as[j][k] == bs[i] ){
                        as[j][k] = 0;
                    }
                }
            }
        }

        // 斜めのチェック
        if( as[0][0] == 0 && as[1][1] == 0 && as[2][2] == 0 ){
            System.out.println( "Yes" );
            return;
        }
        if( as[0][2] == 0 && as[1][1] == 0 && as[2][0] == 0 ){
            System.out.println( "Yes" );
            return;
        }
        for( int i = 0 ; i < 3 ; i++ ){
            if( as[i][0] == 0 && as[i][1] == 0 && as[i][2] == 0 ){
                System.out.println("Yes");
                return;
            }
            if( as[0][i] == 0 && as[1][i] == 0 && as[2][i] == 0 ){
                System.out.println("Yes");
                return;
            }
        }

        System.out.println( "No" );

    }

}