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)

s = gets().split(" ");
for h in 0..2
 s[h] = s[h].to_i;
end

s_a = s.sort;

for h in 0..2
 s_a[h] = s_a[h].to_i;
end

eo = Array.new(2,0);
count = 0;

for i in 0..2
	if s_a[i]%2==0
		eo[0]+=1;
	elsif s_a[i]%2==1
		eo[1]+=1;
	end
end

big = 0;
small = 0;
ant = 0;

if eo[0] ==3
	count += (s_a[2]-s_a[1])/2;
	count += (s_a[2]-s_a[0])/2;
elsif eo[0]==2
	for j in 0..2
		if s_a[j] %2 == 0
			s_a[j] +=1;
		end
	end
	count = (s_a[2]-s_a[1])/2 + (s_a[2]-s_a[0])/2 +1;
elsif eo[0]==1
	for j in 0..2
		if s_a[j]%2 == 1
			s_a[j] += 1;
		end
	end
	count = (s_a[2]-s_a[1])/2 + (s_a[2]-s_a[0])/2 +1;
elsif eo[0]==0
	count += (s_a[2]-s_a[1])/2;
	count += (s_a[2]-s_a[0])/2;
end

puts(count);