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 Ruby (2.3.3) (judge verdict: Wrong Answer)

#!/usr/bin/env ruby

require 'pp'
attrs = Array.new
while line = $stdin.gets
  attrs << line.chomp.split
end

A = Array.new(3).map{Array.new(3,0)}
B = Array.new(3).map{Array.new(3,false)}
A[0] = attrs[0]
A[1] = attrs[1]
A[2] = attrs[2]

flatArray = A.flatten
N = attrs.length - 3
atari = 0
result = false

N.times do |n|
  num = attrs[n+3][0]
  next unless flatArray.include?(num)
  atari += 1
  index = flatArray.index(num) + 1
  i = index / 3
  j = index % 3
  i = 2 if i == 3
  j = 3 if j == 0
  j -= 1
  B[i][j] = true
  next if atari < 3
  # 横検索
  if (B[0].select{|b| b == true}).length == 3 then
    result = true
    break
  end
  if (B[1].select{|b| b == true}).length == 3 then
    result = true
    break
  end
  if (B[2].select{|b| b == true}).length == 3 then
    result = true
    break
  end
  # 縦検索
  if B[0][0] == true && B[1][0] == true && B[2][0] == true then
    result = true
    break
  end
  if B[0][1] == true && B[1][1] == true && B[2][2] == true then
    result = true
    break
  end
  if B[0][2] == true && B[1][2] == true && B[2][2] == true then
    result = true
    break
  end
  # 斜め
  if B[0][0] == true && B[1][1] == true && B[2][2] == true then
    result = true
    break
  end
  if B[0][2] == true && B[1][1] == true && B[2][0] == true then
    result = true
    break
  end

end

puts result ? "Yes" : "No"