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 Python (3.4.3) (judge verdict: Accepted)

import numpy as np
mat = [list(map(int, input().split())) for i in range(3)]
array = np.array(mat, dtype=np.int64)
a20 = array[2]-array[0]
a10 = array[1]-array[0]
b20 = array[:,2]-array[:,0]
b10 = array[:,1]-array[:,0]
flag  = False
if (a20==a20[0]).all() and (a10==a10[0]).all() and (b20==b20[0]).all() and (b10==b10[0]).all():
    for a1 in range(0, 101):
        a2 = a1+ a10[0]
        a3 = a1 + a20[0]
        b1 = array[0,0] - a1
        b2 = b1 + b10[0]
        b3 = b1 +b20[0]
        A = np.array([[a1, a2, a3]]).T
        B = np.array([[b1, b2, b3]])
        C = A+B
        if (C == array).all():
            flag =True
            break
if flag==True:
    print('Yes')
else:
    print('No')