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(map(int,input().split())) for i in range(3)]    # 二重リスト（配列）に標準入力　
n = int(input())                                           # 標準入力（整数）
d = [[0 for i in range(3)] for j in range(3)]              # 要素が全て0の３×３の二重リスト（配列）

for k in range(n):              #  n回繰り返す
    b = int(input())            #　標準入力（整数）
    for i in range(3):          
        for j in range(3):
            if a[i][j] == b:    #  aに、bと同じ数字があるか見る
                d[i][j] = 1     #　あれば、対応するdに1を代入

ans = 'No'                      #  最初はNoと仮定
count = 0

# 横がビンゴか確認
for i in range(3):
    if count == 3:              # １行で3つ１があるか確認
            ans = 'Yes'         # あればビンゴなので'Yes'
            break;              # 内側のループを抜ける
    count = 0
    for j in range(3):
        count += d[i][j]

# 縦がビンゴか確認
count = 0
for i in range(3):
    if ans == 'Yes' or count == 3: 
        break;
    count = 0
    for j in range(3):
        count += d[j][i]
    if count == 3:
        ans = 'Yes'

# 左上から右下にかけて斜めにビンゴしているか確認
count = 0
for i in range(3):
    if ans == 'Yes':
        break;
    count += d[i][i]
    if count == 3:
        ans = 'Yes'

#右上から左下にかけて斜めにビンゴしているか確認
count = 0
for i in range(3):
    if ans == 'Yes':
        break;
    count += d[i][2 - i]
    if count == 3:
        ans = 'Yes'
            

print(ans)