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 Ruby (2.3.3) (judge verdict: Accepted)

def ok(c)
  a=[0]*3
  b=[0]*3
  0.upto(100) do |i|
    a[0] = i
    b[0] = c[0][0] - a[0]
    b[1] = c[0][1] - a[0]
    b[2] = c[0][2] - a[0]
    a[1] = c[1][0] - b[0]
    a[2] = c[2][0] - b[0]
    f=true
    3.times do |i|
      3.times do |j|
        f = false if c[i][j]!=a[i]+b[j]
      end
    end
    return true if f
  end
  false
end
puts ok(3.times.map{gets.split.map &:to_i}) ? :Yes : :No