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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import qualified Data.ByteString.Char8 as BS
import Data.Maybe (fromJust)
import Data.Array.IArray
import Control.Monad

solve :: Array (Int,Int) Int -> [Int]
solve c = do
    a1 <- [0..100]
    let b1 = c!(1,1) - a1
        b2 = c!(1,2) - a1
        b3 = c!(1,3) - a1
        a2 = c!(2,1) - b1
        a3 = c!(3,1) - b1
    guard $  a2+b2 == (c!(2,2))
          && a2+b3 == (c!(2,3))
          && a3+b2 == (c!(3,2))
          && a3+b3 == (c!(3,3))
    return a1

main :: IO ()
main = do
    c <- listArray ((1,1),(3,3))
        . concatMap (map (fst . fromJust . BS.readInt) . BS.words)
        . BS.lines <$> BS.getContents :: IO (Array (Int,Int) Int)
    putStrLn $ if null (solve c) then "No" else "Yes"