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)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let c: Vec<Vec<i32>> = (0..3).map(|_| (0..3).map(|_| read::<i32>()).collect()).collect();
    let a1 = 0;
    let b1 = c[0][0];
    let b2 = c[0][1];
    let b3 = c[0][2];

    let a2 = c[1][0] - b1;
    let a3 = c[2][0] - b1;

    if c[0][0] == (a1 + b1) &&
        c[0][1] == (a1 + b2) &&
        c[0][2] == (a1 + b3) &&
        c[1][0] == (a2 + b1) &&
        c[1][1] == (a2 + b2) &&
        c[1][2] == (a2 + b3) &&
        c[2][0] == (a3 + b1) &&
        c[2][1] == (a3 + b2) &&
        c[2][2] == (a3 + b3) {
            println!("Yes");
        }else{
            println!("No");
        }
    
}