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 Rust (1.15.1) (judge verdict: Accepted)

fn read<T: std::str::FromStr>() -> T {
	let mut s = String::new();
	std::io::stdin().read_line(&mut s).ok();
	s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
	read::<String>().split_whitespace()
		.map(|e| e.parse().ok().unwrap()).collect()
}

fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
	(0..n).map(|_| read_vec()).collect()
}

pub trait LexicalPermutation {
	/// Return `true` if the slice was permuted, `false` if it is already
	/// at the last ordered permutation.
	fn next_permutation(&mut self) -> bool;
}
impl<T> LexicalPermutation for [T] where T: Ord {
	/// Original author in Rust: Thomas Backman <serenity@exscape.org>
	fn next_permutation(&mut self) -> bool {
		// These cases only have 1 permutation each, so we can't do anything.
		if self.len() < 2 { return false; }

		// Step 1: Identify the longest, rightmost weakly decreasing part of the vector
		let mut i = self.len() - 1;
		while i > 0 && self[i-1] >= self[i] {
			i -= 1;
		}

		// If that is the entire vector, this is the last-ordered permutation.
		if i == 0 {
			return false;
		}

		// Step 2: Find the rightmost element larger than the pivot (i-1)
		let mut j = self.len() - 1;
		while j >= i && self[j] <= self[i-1]  {
			j -= 1;
		}

		// Step 3: Swap that element with the pivot
		self.swap(j, i-1);

		// Step 4: Reverse the (previously) weakly decreasing part
		self[i..].reverse();

		true
	}
}

fn main() {
	let bingo: Vec<Vec<usize>> = read_vec2(3);
	let n = read();
	let mut b: Vec<usize> = Vec::new();
	for _ in 0..n {
		b.push(read());
	}

	let mut bingo_result = vec![vec![false; 3]; 3];

	for bi in b {
		for i in 0..3 {
			for j in 0..3 {
				if bingo[i][j] == bi {
					bingo_result[i][j] = true;
				}
			}
		}
	}

	let mut flag = false;
	for i in 0..3 {
		if bingo_result[i][0] && bingo_result[i][1] && bingo_result[i][2] {
			flag = true;
			break;
		}
	}
	for i in 0..3 {
		if bingo_result[0][i] && bingo_result[1][i] && bingo_result[2][i] {
			flag = true;
			break;
		}
	}
	if bingo_result[0][0] && bingo_result[1][1] && bingo_result[2][2] {
		flag = true;
	}
	if bingo_result[2][0] && bingo_result[1][1] && bingo_result[0][2] {
		flag = true;
	}

	if flag {
		println!("Yes");
	} else {
		println!("No");
	}
}