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: Wrong Answer)

n = 3
e = [[int(i) for i in input().split()] for i in range(n)] 

for i in range(101):
    a1 = i
    b1 = e[0][0] - a1
    b2 = e[0][1] - a1
    b3 = e[0][2] - a1
    a2 = e[1][0] - b1
    a3 = e[2][0] - b1

    if b1 < 0:
        break
    if b2 < 0:
        break
    if b3 < 0:
        break
    if a2 < 0:
        break
    if a3 < 0:
        break
    
    if a2+b2 != e[1][1]:
        break
    if a3+b2 != e[2][1]:
        break
    if a2+b3 != e[1][2]:
        break
    if a3+b3 != e[2][2]:
        break
    print("yes")
    exit()
    

print("no")