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 read<T: std::str::FromStr>(
    stdin_lock: &mut std::io::StdinLock,
    buf: &mut Vec<u8>,
    delimiter: u8,
) -> T {
    buf.clear();
    let l = std::io::BufRead::read_until(stdin_lock, delimiter, buf).unwrap();
    buf.truncate(l - 1); // remove delimiter
    let s = unsafe { std::str::from_utf8_unchecked(&buf) };
    s.parse().unwrap_or_else(|_| panic!("read"))
}

fn main() {
    let stdin = std::io::stdin();
    let mut stdin_lock = stdin.lock();
    let mut buf: Vec<u8> = Vec::new();
    let mut cv = vec![vec![0i32; 3]; 3];
    for y in 0..3 {
        for x in 0..3 - 1 {
            cv[y][x] = read(&mut stdin_lock, &mut buf, b' ');
        }
        cv[y][3 - 1] = read(&mut stdin_lock, &mut buf, b'\n');
    }

    let mut ans = false;
    for a1 in -100..100 + 1 {
        let b1 = cv[0][0] - a1;
        let b2 = cv[0][1] - a1;
        let b3 = cv[0][2] - a1;
        let a2 = cv[1][0] - b1;
        let a3 = cv[2][0] - b1;
        if a2 == cv[1][1] - b2 && a2 == cv[1][2] - b3 && a3 == cv[2][1] - b2 && a3 == cv[2][2] - b3
        {
            ans = true;
            break;
        }
    }
    println!("{}", if ans { "Yes" } else { "No" });
}