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: Accepted)

# import numpy as np
# A=np.zeros((3,3))
# A[0][:]=[int(i) for i in input().split()]
# A[1][:]=[int(i) for i in input().split()]
# A[2][:]=[int(i) for i in input().split()]
# N=int(input())
# B=np.zeros((N,1))
# B[:][0]=[int(i) for i in input().splitlines()]
# print(B)

# # A=[]
# # for i in range(3)

A=[]
answer=[[0,0,0],[0,0,0],[0,0,0]]
for i in range(3):
    A.append(list(map(int,input().split())))
N=int(input())
B=[int(input()) for i in range(N)]
for j in range(N):
    for k in range(3):
        for l in range(3):
            if B[j]==A[k][l]:
                answer[k][l]=1
i=0
J=0
for i in range(3):
    if answer[i][0]==answer[i][1]==answer[i][2]==1:
        print("Yes")
        J=J+1
        break
    elif answer[0][i]==answer[1][i]==answer[2][i]==1:
        J=J+1
        print("Yes")
        break
if J==0 and answer[0][0] == answer[1][1] == answer[2][2] == 1:
    J=J+1
    print("Yes")
if J == 0 and answer[0][2] == answer[1][1] == answer[2][0]== 1:
    J=J+1
    print("Yes")
if J == 0:
    print("No")