Problem:
Score: 300 points

Problem Statement

We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j.

Determine if he is correct.

Constraints

c_{i, j} \ (1 \leq i \leq 3, 1 \leq j \leq 3) is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

c_{1,1} c_{1,2} c_{1,3}
c_{2,1} c_{2,2} c_{2,3}
c_{3,1} c_{3,2} c_{3,3}

Output

If Takahashi's statement is correct, print Yes; otherwise, print No.

Sample Input 1

1 0 1
2 1 2
1 0 1

Sample Output 1

Yes

Takahashi is correct, since there are possible sets of integers such as: a_1=0,a_2=1,a_3=0,b_1=1,b_2=0,b_3=1.

Sample Input 2

2 2 2
2 1 2
2 2 2

Sample Output 2

No

Takahashi is incorrect in this case.

Sample Input 3

0 8 8
0 8 8
0 8 8

Sample Output 3

Yes

Sample Input 4

1 8 6
2 9 7
0 7 7

Sample Output 4

No

Solution in Rust (1.15.1) (judge verdict: Accepted)

#![allow(unused_macros, unused_imports, dead_code)]

use std::io;

fn read_line() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).unwrap();
    s
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    io::stdin().read_line(&mut s).ok();
    s.trim().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

macro_rules! from_line {
    ($($a:ident : $t:ty),+) => {
        $(let $a: $t;)+
        {
            let _line = read_line();
            let mut _it = _line.trim().split_whitespace();
            $($a = _it.next().unwrap().parse().unwrap();)+
            assert!(_it.next().is_none());
        }
    };
}

fn main() {
    let C: Vec<Vec<i32>> = (0..3).map(|_| read_vec()).collect();

    let mut ok: bool = false;
    for i in 0..101 {
        for j in 0..101 {
            for k in 0..101 {
                if C[0][0] - i == C[0][1] - j && C[0][0] - i == C[0][2] - k {
                    if C[1][0] - i == C[1][1] - j && C[1][0] - i == C[1][2] - k {
                        if C[2][0] - i == C[2][1] - j && C[2][0] - i == C[2][2] - k {
                            ok = true;
                        }
                    }
                }
            }
        }
    }

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