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()
}

fn main() {
    let a: Vec<Vec<i32>> = read_vec2(3);
    let n: i32 = read();
    let b: Vec<i32> = (0..n).map(|_| read()).collect();

    let mut table = [[false; 3]; 3];
    let mut yes = false;

    for (i, row) in a.iter().enumerate() {
        for (j, col) in row.iter().enumerate() {
            for v in &b {
                if v == col {
                    table[i][j] = true;
                }
            }
        }
    }

    // yoko
    for i in table.iter() {
        if i.iter().all(|&x| x) {
            yes = true;
            break;
        }
    }

    // tate
    if yes != true {
        if table[0][0] && table[1][0] && table[2][0] ||
        table[0][1] && table[1][1] && table[2][1] ||
        table[0][2] && table[1][2] && table[2][2] {
            yes = true;
        }
    }

    // naname
    if yes != true {
        if table[0][0] && table[1][1] && table[2][2] {
            yes = true;
        }

        if table[0][2] && table[1][1] && table[2][0] {
            yes = true;
        }
    }

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