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)

def inlt():
    return(list(map(int, input().split())))


def inp():
    return(int(input()))


A = []
for t in range(3):
    A.append(inlt())

n = inp()
for t in range(n):
    b = inp()
    # mark the position of b
    for i in range(3):
        for j in range(3):
            if A[i][j] == b:
                A[i][j] = 0
res = "No"
# see row wise:
for i in range(3):
    s = 0
    for j in range(3):
        s += A[i][j]
    if s == 0:
        res = "Yes"
        break
# see column wise
for i in range(3):
    s = 0
    for j in range(3):
        s += A[j][i]
    if s == 0:
        res = "Yes"
        break
# see diagonal
s = 0
for i in range(3):
    for j in range(3):
        if i == j:
            s += A[j][i]
if s == 0:
    res = "Yes"

s = 0
for i in range(3):
    for j in range(3):
        if i == j:
            s += A[j][i]
if s == 0:
    res = "Yes"

if A[0][2] == A[1][1] == A[2][0]:
    res = "Yes"

print(res)