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 C++ (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#ifdef _WIN32
#include "debug.hpp"
#endif
using namespace std;

#define rep(i, N)  for(int i = 0; i < (N); i++)
#define reps(i, N) for(int i = 1; i <= (N); i++)
#define repr(i, N) for(int i = (N) - 1; i >= 0; i--)
#define pub push_back

template<typename T> 
void chmax(T &a, T b){ a = max(a, b); };
template<typename T> 
void chmin(T &a, T b){ a = min(a, b); };

typedef long long ll;
typedef pair<int, int> P;
const int INF = 100000000;
const ll LINF = 10000000000000000;
const int MOD = 1000000007;
const int dx[9] = { 0, 1,  0, -1, 1,  1, -1, -1, 0};
const int dy[9] = { 1, 0, -1,  0, 1, -1, -1,  1, 0};
//--------------------------------------//
int A, B, C;

int c(int a, int b){
	if(a > b)
		return a - b;
	return (b - a + 1) / 2 + (b - a) % 2;
}
void solve(){
	int d[3] = {
		c(A, C) + c(B, C),
		c(C, A) + c(B, A),
		c(C, B) + c(A, B),
	};
	cout << min(d[0], min(d[1], d[2])) << endl;

}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> A >> B >> C;

		solve();
	return 0;
}