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 Python (3.4.3) (judge verdict: Wrong Answer)

def main():
    a = [list(map(int, input().split())) for _ in range(3)]
    n = int(input())
    b = [int(input()) for _ in range(n)]
    card = [[0]*3]*3
    for i in range(3): #aを順番に見ていく
        for j in range(3):
            if a[i][j] in b:
                card[i][j] =1
    for i in range(3): # card
        if i == 0:
            if card[i][i] == card[i][i+1] == card[i][i+2]==1:#縦の探索
                print("Yes")
                break
            elif card[i][i] == card[i+1][i] == card[i+2][i]==1:#横の探索
                print("Yes")
                break
            elif card[i][i] == card[i+1][i+1] == card[i+2][i+2]==1:#0の時の斜め            
                print("Yes")
                break
        elif i == 1:
            if card[i-1][i] == card[i][i] == card[i+1][i]==1:
                print("Yes")
                break
        elif i == 2:
            if card[i-2][i] == card[i-1][i-1] == card[i][i-2] == 1:#2のときの斜
                print("Yes") 
                break   
    else:print("No")

if __name__ == "__main__":
    main()