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 (3.8.2) (judge verdict: Accepted)

l = []
for i in range(3):
    c1,c2,c3 = map(int, input().split())
    l.append([c1,c2,c3])

# assign a1 as 0
b1 = l[0][0]
b2 = l[1][0]
b3 = l[2][0]

flag = 0
for i in range(3):
    if not (l[0][1]-b1 == l[1][1]-b2 == l[2][1]-b3):
        flag = 1
    if not (l[0][2]-b1 == l[1][2]-b2 == l[2][2]-b3):
        flag = 1
if flag == 1:
    print('No' , flush=True)
else:
    print('Yes' , flush=True)

# TLE
# l = []
# for i in range(3):
#     c1,c2,c3 = map(int, input().split())
#     l.append([c1,c2,c3])

# def check_sum(i,j):
#     for r in range(3):
#         if r == i:
#             continue;
#         for c in range(3):
#             if c == j:
#                 continue;
#             sum_a = l[i][j] + l[r][c]
#             sum_b = l[r][j] + l[i][c]
#             if sum_a == sum_b:
#                 return 1
#             else:
#                 return -1
#                 break;

# for i in range(3):
#     for j in range(3):
#         flag = check_sum(i, j)
#         if flag == -1:
#             break;

# if flag == 1:
#     print('Yes' , flush=True)
# else:
#     print('No' , flush=True)