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 PyPy3 (2.4.0) (judge verdict: Accepted)

import copy
def solve():
    c = []
    for i in range(3):
        tmp = list(map(int,input().split()))
        c.append(tmp)
    
    for a1 in range(0, 100+1):
        b1 = c[0][0] - a1
        b2 = c[0][1] - a1
        b3 = c[0][2] - a1

        a2 = c[1][0] - b1
        if a2 != c[1][1] - b2 or a2 != c[1][2] - b3:
            continue


        a3 = c[2][0] - b1
        if a3 != c[2][1] - b2 or a3 != c[2][2] - b3:
            continue

        print('Yes')
        break
    else:
        print('No')

if __name__ == '__main__':
    solve()