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)

a = []
3.times do
  a << gets.chomp.split(" ").map{|i|i = i.to_i}
end
n = gets.chomp.to_i
b = []
n.times do
  b << gets.chomp.to_i
end
bingo = Array.new(3, Array.new(3, 0) )

b.each do |bElm|
  3.times do|i|
    3.times do|j|
      if a[i][j] == bElm then
        bingo[i][j] = 1
      end
    end
  end
end

# 横を調べる
if ((bingo[0][0] == 1 && bingo[0][1] == 1 && bingo[0][2]) == 1 || 
    (bingo[1][0] == 1 && bingo[1][1] == 1 && bingo[1][2] == 1) || 
    (bingo[2][0] == 1 && bingo[2][1] == 1 && bingo[2][2] == 1)) then
  puts "Yes"
# 縦を調べる
elsif ((bingo[0][0] == 1 && bingo[1][0] == 1 && bingo[2][0] == 1) || 
       (bingo[0][1] == 1 && bingo[1][1] == 1 && bingo[2][1] == 1) || 
       (bingo[0][2] == 1 && bingo[1][2] == 1 && bingo[2][2] == 1)) then
  puts "Yes"
# 斜めを調べる
elsif ((bingo[0][0] == 1 && bingo[1][1] == 1 && bingo[2][2] == 1) || 
       (bingo[2][0] == 1 && bingo[1][1] == 1 && bingo[0][2] == 1)) then
  puts "Yes"
else
  puts "No"
end