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::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

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

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let mut vec: Vec<Vec<usize>> = Vec::new();
    for _i in 0..3 {
        let mut v = Vec::new();
        for _j in 0..3 {
            v.push(iterator.next().unwrap().parse().unwrap());
        }
        vec.push(v);
    }

    let n: usize = iterator.next().unwrap().parse().unwrap();
    for _i in 0..n {
        let b: usize = iterator.next().unwrap().parse().unwrap();
        for j in 0..3 {
            for k in 0..3 {
                if vec[j][k] == b {
                    vec[j][k] =0;
                }
            }
        }
    }

    return if (vec[0][0] == 0 && vec[0][1] == 0 && vec[0][2] == 0)
        || (vec[1][0] == 0 && vec[1][1] == 0 && vec[2][1] == 0)
        || (vec[2][0] == 0 && vec[2][1] == 0 && vec[2][2] == 0)
        || (vec[0][0] == 0 && vec[1][0] == 0 && vec[2][0] == 0)
        || (vec[0][1] == 0 && vec[1][1] == 0 && vec[2][1] == 0)
        || (vec[0][2] == 0 && vec[1][2] == 0 && vec[2][2] == 0)
        || (vec[0][0] == 0 && vec[1][1] == 0 && vec[2][2] == 0)
        || (vec[0][2] == 0 && vec[1][1] == 0 && vec[2][0] == 0) {
        "Yes".to_string()
    } else {
        "No".to_string()
    }
}