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)

use std::io::{self, BufRead};

fn main() {
    println!("{}", if solve() { "Yes" } else { "No" });
}

fn solve() -> bool {
    let mut r = io::BufReader::new(io::stdin());

    let mut buf = String::new();

    let mut a: Vec<Vec<usize>> = Vec::with_capacity(3);
    for _ in 0..3 {
        buf.clear();
        let _ = r.read_line(&mut buf).unwrap();
        a.push(buf.split(" ").map(|a| a.trim().parse().unwrap()).collect());
    }

    let _ = {
        buf.clear();
        r.read_line(&mut buf).unwrap()
    };

    let mut bs = [false; 101];
    for i in r.lines().map(|a| a.unwrap().trim().parse::<usize>().unwrap()) {
        bs[i] = true;
    }

    for i in 0..3 {
        if bs[a[i][0]] && bs[a[i][1]] && bs[a[i][2]] {
            return true;
        }
    }

    for j in 0..3 {
        if bs[a[0][j]] && bs[a[1][j]] && bs[a[2][j]] {
            return true;
        }
    }

    if bs[a[0][0]] && bs[a[1][1]] && bs[a[2][2]] {
        return true;
    }

    if bs[a[0][2]] && bs[a[1][1]] && bs[a[2][0]] {
        return true;
    }

    return false;
}