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 (3.8.2) (judge verdict: Wrong Answer)

import bisect,collections,copy,heapq,itertools,math,numpy,string
import sys
sys.setrecursionlimit(10**7)
 
def _S(): return sys.stdin.readline().rstrip()
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def LS(): return list(sys.stdin.readline().rstrip().split())
 
def main():
    A = [LI() for _ in range(3)]
    N = I()
    B = [I() for _ in range(N)]
    # print(B) 
    
    for i in range(3):
        for j in range(3):
            for k in range(N):
                if A[i][j]== B[k]:
                    A[i][j] = 0
                    break

    ans = 'No'
    for i in range(3):
            if (A[i][0]==0 and A[i][1]==0 and A[i][2]==0) or (A[0][i]==0 and A[1][i]==0 and A[2][i]==0):
              ans = 'Yes'
              break  
    if (A[0][2]==0 and A[1][1]==0 and A[2][0]==0) or A[0][0]==0 and A[1][1]==0 and A[2][2]:
        ans = 'Yes'
    print(ans)
main()