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() {
    let mut v: Vec<Vec<usize>> = vec![Vec::new(); 3];
    for i in 0..v.len() {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        v[i] = s.trim().split(' ').map(|x| x.parse().unwrap()).collect();
    }

    for i in 0..v[0][0]+1 {
        for j in 0..v[1][1]+1 {
            for k in 0..v[2][2]+1 {
                let x = v[0][0] - i;
                let y = v[1][1] - j;
                let z = v[2][2] - k;

                if v[0][1] == i + y && v[0][2] == i + z &&
                   v[1][0] == j + x && v[1][2] == j + z &&
                   v[2][0] == k + x && v[2][1] == k + y {
                       println!("Yes"); return;
                   }
            }
        }
    }

    println!("No");
}