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)

fn main() {
    // cells[j][i] == a[i] + b[j]
    let c = read_lines(3);
    let mut a = vec![0, 0, 0];
    let mut b = vec![0, 0, 0];

    a[0] = 0;
    b[0] = c[0][0] - a[0];
    b[1] = c[1][0] - a[0];
    b[2] = c[2][0] - a[0];
    a[1] = c[0][1] - b[0];
    a[2] = c[0][2] - b[0];

    for j in 0..3 {
        for i in 0..3 {
            if c[j][i] != a[i] + b[j] {
                println!("No");
                return;
            }
        }
    }
    println!("Yes");
}

fn read_lines(n: i32) -> Vec<Vec<i32>> {
    let mut v2 = Vec::new();
    for _ in 0..n {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).ok();
        let v = s.trim()
            .split_whitespace()
            .map(|e| e.parse().ok().unwrap())
            .collect();
        v2.push(v);
    }
    v2
}