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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <cmath>
#include <iostream>
#include <sstream>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>

#ifdef _DEBUG
#define DMP(x) cerr << #x << ": " << x << "\n"
#else
#define DMP(x) ((void)0)
#endif

const int MOD = 1000000007, INF = 1111111111; 
using namespace std;
using lint = long long;

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	vector<pair<int,int>> A(3);
	for (int i = 0; i < 3; i++) {
		cin >> A[i].first;
		A[i].second = A[i].first % 2;
	}

	sort(A.begin(), A.end());

	int ans = 0;
	if (A[0].second != A[1].second || A[1].second != A[2].second) {
		ans++;
		for (int i = 0; i < 3; i++) {
			if (A[i].second == A[(i + 1) % 3].second) {
				A[i].first++, A[(i + 1) % 3].first++;
				break;
			}
		}

	}
	ans += (A[2].first * 2 - A[1].first - A[0].first) / 2;
	cout << ans << "\n";

	return 0;
}