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)

A_list_yoko1 = list(map(int, input().split()))
A_list_yoko2 = list(map(int, input().split()))
A_list_yoko3 = list(map(int, input().split()))
N = int(input())
B_list = []
for i in range(N):
    B_list.append(int(input()))

A_list_tate1 = []
A_list_tate2 = []
A_list_tate3 = []
A_list_naname1 = []
A_list_naname2 = []

A_list_tate1.append(A_list_yoko1[0])
A_list_tate1.append(A_list_yoko2[0])
A_list_tate1.append(A_list_yoko3[0])
A_list_tate2.append(A_list_yoko1[1])
A_list_tate2.append(A_list_yoko2[1])
A_list_tate2.append(A_list_yoko3[1])
A_list_tate3.append(A_list_yoko1[2])
A_list_tate3.append(A_list_yoko2[2])
A_list_tate3.append(A_list_yoko3[2])
A_list_naname1.append(A_list_yoko1[0])
A_list_naname1.append(A_list_yoko2[1])
A_list_naname1.append(A_list_yoko3[2])
A_list_naname2.append(A_list_yoko1[2])
A_list_naname2.append(A_list_yoko2[1])
A_list_naname2.append(A_list_yoko3[0])

A_list_yoko1_c = []
A_list_yoko2_c = []
A_list_yoko3_c = []
A_list_tate1_c = []
A_list_tate2_c = []
A_list_tate3_c = []
A_list_naname1_c = []
A_list_naname2_c = []
for i in B_list:
    if i in A_list_yoko1:
        A_list_yoko1_c.append(A_list_yoko1.index(i))
    if i in A_list_yoko2:
        A_list_yoko2_c.append(A_list_yoko2.index(i))
    if i in A_list_yoko3:
        A_list_yoko3_c.append(A_list_yoko3.index(i))
    if i in A_list_tate1:
        A_list_tate1_c.append(A_list_tate1.index(i))
    if i in A_list_tate2:
        A_list_tate2_c.append(A_list_tate2.index(i))
    if i in A_list_tate3:
        A_list_tate3_c.append(A_list_tate3.index(i))
    if i in A_list_naname1:
        A_list_naname1_c.append(A_list_naname1.index(i))
    if i in A_list_naname2:
        A_list_naname2_c.append(A_list_naname2.index(i))

if list(A_list_naname1_c) == 3 or list(A_list_naname2_c) == 3 or list(A_list_tate1_c) == 3 or list(A_list_tate2_c) == 3 or list(A_list_tate3_c) == 3 or list(A_list_yoko1_c) == 3 or list(A_list_yoko2_c) == 3 or list(A_list_yoko3_c) == 3:
    print("Yes")
else:
    print("No")