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: Wrong Answer)

use std::io;

fn main() {
    let stdin = io::stdin();
    let mut buf = String::new();

    let a: Vec<_> = (0..3)
        .map(|_| {
            buf.clear();
            stdin.read_line(&mut buf).unwrap();
            buf.split_whitespace()
                .map(|x| x.parse().unwrap())
                .collect::<Vec<u32>>()
        })
        .collect();
    buf.clear();
    stdin.read_line(&mut buf).unwrap();
    let n: u32 = buf.trim_right().parse().unwrap();
    let mut hori = [0; 3];
    let mut vert = [0; 3];
    let mut diag = [0; 2];
    for _ in 0..n {
        buf.clear();
        stdin.read_line(&mut buf).unwrap();
        let x: u32 = buf.trim_right().parse().unwrap();

        for i in 0..3 {
            for j in 0..3 {
                if x == a[i][j] {
                    hori[i] += 1;
                    vert[j] += 1;
                    if i == j {
                        diag[0] += 1;
                    }
                    if i == 3 - j {
                        diag[1] += 1;
                    }
                }
            }
        }
    }
    println!(
        "{}",
        if hori
            .iter()
            .chain(vert.iter())
            .chain(diag.iter())
            .any(|x| x == &3)
        {
            "Yes"
        } else {
            "No"
        }
    );
}