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 <iostream>
#include <sstream>
#include <regex>
#include <cstdio>
#include <cmath>
#include <cctype>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
#include <iomanip>

using namespace std;
using ll = long long int;
#define rep(i,n) for(int i = 0; i < n; i++)
#define FOR(i, a, b)  for(int i = (a); i < (b) ; i++)
#define pb push_back
#define SORT(v,n) sort(v, v+n)
#define ALL(x) (x).begin(),(x).end()
#define debug(x) cerr << #x << ": " << x << '\n'
#define elif else if
#define int ll
//const int INF = 100100100;
const int INF = (1LL << 32);
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
int dx[8] = { 1, 0, -1, 0, 1, -1, -1, 1 };
int dy[8] = { 0, 1, 0, -1, 1, 1, -1, -1 };
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;

signed main() {
	ios::sync_with_stdio(false);
	int a, b, c;
	cin >> a >> b >> c;
	int x = max(a,max(b,c));
	int m = 0;
	if ((a + b + c) % 2 != 3 * x % 2) {
		m = x + 1;
	}
	else {
		m = x;
	}
	cout << ((3 * m) - (a + b + c)) / 2 << endl;
	return 0;
}