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: Accepted)

A, B, C = STDIN.gets.chomp.split(' ').map(&:to_i)

odd = 0
even = 0

min = [A, B, C].sort[0]

a = A - min
b = B - min
c = C - min

[a, b, c].each do |n|
  if n % 2 == 0
    even += 1
  else
    odd += 1
  end
end

arr = [a, b, c].sort.reverse



cnt = 0
# puts "#{cnt} #{arr.join(',')}"

while arr[1] - arr[2] > 2
  arr[0] -= 2
  arr[1] -= 2
  cnt += 1
  # puts "#{cnt} #{arr.join(',')}"
end

while arr[0] - arr[1] > 2
  arr[0] -= 2
  cnt += 2
  # puts "#{cnt} #{arr.join(',')}"
end


case [arr[0] - arr[1], arr[1] - arr[2]]
when [1, 1]
  # 3, 2, 1
  cnt += 3
when [2, 2]
  # 4, 2, 0
  cnt += 3
when [0, 2]
  # 2, 2, 0
  cnt += 1
when [2, 0]
  # 2, 0, 0
  cnt += 2
when [1, 2]
  # 3, 2, 0
  cnt += 2
when [2, 1]
  # 3, 1, 0
  cnt += 4
when [0, 2]
  # 3, 3, 1
  cnt += 1
when [2, 0]
  # 3, 1, 1
  cnt += 2
when [1, 0]
  # 2, 1, 1
  cnt += 1
when [0, 1]
  # 1, 1, 0
  cnt += 2
when [0, 0]
  #pass
end

puts cnt