Problem:
Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE OverloadedLists #-}

module Main where

import qualified Data.Text as T
import qualified Data.Text.IO as T
import Data.List
import Control.Applicative
import Control.Monad
import Data.Char

main :: IO ()
main = do
  [a, b, c] <- map read . words <$> getLine
  print $ solve a b c

solve :: Int -> Int -> Int -> Int
solve a b c = (\(_, _, _, res) -> res)
  . thirdProcess
  . secondProcess
  . firstProcess $ sorted (a, b, c, 0) where
  sorted :: (Int, Int, Int, Int) -> (Int, Int, Int, Int)
  sorted (a', b', c', res) = (\[a'', b'', c''] -> (a'', b'', c'', res)) $ sub a' b' c' where
    sub :: Int -> Int -> Int -> [Int]
    sub a''' b''' c''' = sort [a''', b''', c''']
  firstProcess :: (Int, Int, Int, Int) -> (Int, Int, Int, Int)
  firstProcess (a', b', c', res)
    | b' == c'              = (a', b', c', res)
    | otherwise             = firstProcess (a' + 1, b' + 1, c', res + 1)
  secondProcess :: (Int, Int, Int, Int) -> (Int, Int, Int, Int) 
  secondProcess (a', b', c', res)
    | a' >= b'              = (a', b', c', res)
    | otherwise             = secondProcess (a' + 2, b', c', res + 1)
  thirdProcess :: (Int, Int, Int, Int) -> (Int, Int, Int, Int)
  thirdProcess (a', b', c', res)
    | a' == b'              = (a', b', c', res)
    | otherwise             = (a', b' + 1, c' + 1, res + 1)