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)

l1 = gets.chomp.split(" ").map(&:to_i);
l2 = gets.chomp.split(" ").map(&:to_i);
l3 = gets.chomp.split(" ").map(&:to_i);

flatten = l1.concat(l2).concat(l3)

n =  gets.chomp.to_i
n.times do
  b =  gets.chomp.to_i
  flatten[flatten.index(b)] =true if flatten.include?(b)
end

l1=flatten[0..2]
l2=flatten[3..5]
l3=flatten[6..8]

pp l1
pp l2
pp l3

if (l1[0] == l2[0] && l2[0] == l3[0]) ||
   (l1[1] == l2[1] && l2[1] == l3[1]) ||
   (l1[2] == l2[2] && l2[2] == l3[2]) ||
   (l1[0] == l1[1] && l1[1] == l1[2]) ||
   (l2[0] == l2[1] && l2[1] == l2[2]) ||
   (l3[0] == l3[1] && l3[1] == l3[2]) ||
   (l1[0] == l2[1] && l2[1] == l3[2]) ||
   (l1[2] == l2[1] && l2[1] == l3[0])
puts "Yes"
else
puts "No"
end