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 (1.42.0) (judge verdict: Accepted)

use proconio::input;

#[allow(unused_variables)]
fn main() {
    input! {
        aaa: [[usize; 3]; 3],
        n: usize,
        ba: [usize; n]
    };

    let bingo = aaa
        .into_iter()
        .map(|x| {
            x.iter()
                .map(|y| if ba.contains(y) { true } else { false })
                .collect::<Vec<bool>>()
        })
        .collect::<Vec<Vec<bool>>>();

    let mut ans = "No";

    for i in 0..3 {
        // 縦
        if bingo[0][i] && bingo[1][i] && bingo[2][i] {
            ans = "Yes"
        }

        // 横
        if bingo[i][0] && bingo[i][1] && bingo[i][2] {
            ans = "Yes"
        }
    }

    // 斜め
    if bingo[0][0] && bingo[1][1] && bingo[2][2] {
        ans = "Yes"
    }

    if bingo[0][2] && bingo[1][1] && bingo[2][0] {
        ans = "Yes"
    }

    println!("{}", ans);
}