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)

#![allow(non_snake_case, unused_variables, unused_mut, dead_code)]
//use permutohedron::LexicalPermutation as _;
//use proconio::input;
//use proconio::marker::Chars;
//use proconio::marker::Usize1;
//use std::collections::HashSet;
//use std::collections::HashMap;
//use std::collections::BTreeMap;
//use std::cmp::*;
//use std::num;

fn main() {
    proconio::input! {
        mut A:[usize;9],
        N:usize,
        B:[usize;N]
    }
    for i in 0..N {
        A = A.iter().map(|&a| if a == B[i] { 0 } else { a }).collect();
    }
    //println!("{:?}", A);
    for i in 0..3 {
        if (A[i + 0] == 0 && A[i + 3] == 0 && A[i + 6] == 0)
            || (A[i * 3] == 0 && A[i * 3 + 1] == 0 && A[i * 3 + 2] == 0)
        {
            println!("Yes");
            return;
        }
    }

    if (A[0] == 0 && A[4] == 0 && A[8] == 0) || (A[2] == 0 && A[4] == 0 && A[6] == 0) {
        println!("Yes");
        return;
    }
    println!("No");
}