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)

a1 = list(map(int, input().split()))
a2 = list(map(int, input().split()))
a3 = list(map(int, input().split()))
n = int(input())
s = list(map(int, [input() for i in range(n)]))
l = [
  [0, 0, 0]
  , [0, 0, 0]
  , [0, 0, 0]
]
for i in range(0, 3):
  for j in range(0, n):
    if a1[i] == s[j]:
      l[0][i] = i+1
for i in range(0, 3):
  for j in range(0, n):
    if a2[i] == s[j]:
      l[1][i] = i+1
for i in range(0, 3):
  for j in range(0, n):
    if a3[i] == s[j]:
      l[2][i] = i+1
n = []
for i in l:
  for j in i:
    if j != 0:
      n.append(j)
count = 0
for i in range(0, 3):
  if l[i][0] and l[i][1] and l[i][2]:
    count += 1
    print('Yes')
if (l[0][0] and l[1][1] and l[2][2]) or (l[2][0] and l[1][1] and l[0][2]):
  count += 1
  print('Yes')
else:
  for i in range(1, 4):
    if n.count(i) == 3:
      count += 1
      print('Yes')
if count == 0:
  print('No')