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 Ruby (2.3.3) (judge verdict: Wrong Answer)

arr = gets.split.map(&:to_i) 
arr.sort!

if arr[0]%2 == 0 && arr[1]%2 == 0 && arr[2]%2 == 0 ||  arr[0]%2 == 1 && arr[1]%2 == 1 && arr[2]%2 == 1
	puts (arr[2]-arr[0])/2 + (arr[2]-arr[1])/2 
elsif arr[0]%2 == 0 && arr[1]%2 == 0 || arr[0]%2 == 1 && arr[1]%2 == 1
	puts (arr[2]-arr[0]-1)/2 + (arr[2]-arr[1]-1)/2 + 1
elsif arr[0]%2 == 0 && arr[2]%2 == 0 || arr[0]%2 == 1 && arr[2]%2 == 1
	puts (arr[2]-arr[0])/2 + (arr[2]-arr[1]+1)/2 + 1
else
	puts (arr[2]+1-arr[0])/2 + (arr[2]-arr[1]-1)/2 + 1
end