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::char;
use std::io::Read;

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

    std::io::stdin().read_to_string(&mut buf).unwrap();

    let mut iter = buf.split_whitespace();
    let mut a = vec![vec![0; 3]; 3];
    for i in 0..3 {
        for j in 0..3 {
            a[i][j] = iter.next().unwrap().parse().unwrap();
        }
    }
    let n: usize = iter.next().unwrap().parse().unwrap();
    let b: Vec<i32> = (0..n).map(|_| iter.next().unwrap().parse().unwrap()).collect();

    for i in 0..3 {
        for j in 0..3 {
            a[i][j] = if b.contains(&a[i][j]) { 1 } else { 0 };
        }
    }
    let mut ans = false;
    if a[0][0] + a[0][1] + a[0][2] == 3 { ans = true ;}
    if a[1][0] + a[1][1] + a[1][2] == 3 { ans = true ;}
    if a[2][0] + a[2][1] + a[2][2] == 3 { ans = true ;}
    if a[0][0] + a[1][0] + a[2][0] == 3 { ans = true ;}
    if a[0][1] + a[1][1] + a[2][1] == 3 { ans = true ;}
    if a[0][2] + a[1][2] + a[2][2] == 3 { ans = true ;}
    if a[0][0] + a[1][1] + a[2][2] == 3 { ans = true ;}
    if a[2][0] + a[1][1] + a[0][2] == 3 { ans = true ;}
    
    println!("{}", if ans { "Yes" } else { "No "});
}