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: Runtime Error)

card = Array.new(3) { Array.new(3) }
bingo = Array.new(3) { Array.new(3, false) }

3.times do |i|
  a1, a2, a3 = gets.split.map(&:to_i)
  card[i][0] = a1
  card[i][1] = a2
  card[i][2] = a3
end

# p card

N = gets.to_i

N.times do |i|
  b = gets.to_i
  card.each_with_index do |c, i|
    c.each_with_index do |a, j|
      if a == b
        bingo[i][j] = true
      end
    end
  end
end

# p bingo

ok = false
3.times do |i|
  is_bingo = bingo[i][0] == true && bingo[i][1] == true && bingo[i][2] == true
  if is_bingo
    ok = true
    break
  end
end

if ok
  puts "Yes"
  return
end

3.times do |i|
  is_bingo = bingo[0][i] == true && bingo[1][i] == true && bingo[2][i] == true
  if is_bingo
    ok = true
    break
  end
end

if ok
  puts "Yes"
  return
end

ok = bingo[0][0] == true && bingo[1][1] == true && bingo[2][2] == true
if ok
  puts "Yes"
  return
end

ok = bingo[0][2] == true && bingo[1][1] == true && bingo[2][0] == true


if ok
  puts "Yes"
else
  puts "No"
end