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)

grid = 3.times.map { gets.split.map(&:to_i) }

0.upto(grid[0][0]) do |b1|
  a1 = grid[0][0] - b1

  0.upto(grid[1][1]) do |b2|
    a2 = grid[1][1] - b2

    next if a1 + b2 != grid[0][1]
    next if a2 + b1 != grid[1][0]

    0.upto(grid[2][2]) do |b3|
      a3 = grid[2][2] - b3

      next if a1 + b3 != grid[0][2]
      next if a2 + b3 != grid[1][2]
      next if a3 + b1 != grid[2][0]
      next if a3 + b2 != grid[2][1]

      puts 'Yes'
      exit
    end
  end
end

puts 'No'