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.*;
	class Main{
		public static void main(String args[]){
			Scanner sc = new Scanner(System.in);
			int[][] a = new int[3][3];
			for (int i = 0; i < 3; i++)
				for (int j = 0; j < 3; j++)
					a[i][j] = sc.nextInt();
			int n = sc.nextInt();
			Set<Integer> set = new HashSet<>();
			for (int i = 0; i < n; i++)
				set.add(sc.nextInt());
			int count = 0, countr = 0, countd = 0;
			for (int i = 0; i < 3; i++){
				count = 0;
				for (int j = 0; j < 3; j++){
					//~ System.out.println(a[i] + " 2 " );
					if(set.contains(a[i][j])){
						count++;
						//~ System.out.println(a[i][j] + " 1 ");
					} 
					if(i == j){
							if (set.contains(a[i][j])){
									countd++;
									//~ System.out.println(a[i][j] + " 2 " );
						}
					}
				}
				countr = Math.max(count, countr);
			}
			int countc = 0;
				for (int i = 0; i < 3; i++){
				count = 0;
				for (int j = 0; j < 3; j++){
					if(set.contains(a[j][i]) ){
						count++;
					}
				}
				countc = Math.max(countc, count);
				} 
				
				int countdd = 0;
			if(set.contains(a[0][2]))
				countdd++;
			if(set.contains(a[1][1]))
				countdd++;
			if(set.contains(a[2][0]))
				countdd++;
				
			if(countc == 3 || countr == 3 | countd == 3 || countdd == 3)
				System.out.println("Yes");
			else 
				System.out.println("No");
		}
	}