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)

cij =[]
for i in range(3):
    cij.append(list(map(int, input().split(" "))))

bi_list = []
for b1 in range(-50000, 50000):
    b2 = b1 - (cij[0][0] - cij[0][1])
    b3 = b1 - (cij[0][0] - cij[0][2])

    a1 = cij[0][0]+cij[0][1]+cij[0][2] -b1-b2-b3
    a2 = cij[1][0]+cij[1][1]+cij[1][2] -b1-b2-b3
    a3 = cij[2][0]+cij[2][1]+cij[2][2] -b1-b2-b3
    if a1/3+b1 ==cij[0][0] and a1/3+b2 ==cij[0][1] and a1/3+b3 ==cij[0][2] and a2/3+b1 ==cij[1][0] and a2/3+b2 ==cij[1][1] and a2/3+b3 ==cij[1][2] and a3/3+b1 ==cij[2][0] and a3/3+b2 ==cij[2][1] and a3/3+b3 ==cij[2][2]:
            print("Yes")
            exit()

print("No")