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 <bits/stdc++.h>
#define exrep(i, a, b) for(long long i = a; i <= b; i++)
#define rep(i,n)for(long long i=0;i<(long long)(n);i++)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> P;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vector<ll> > vvl;
typedef vector<vector<P> > vvp;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
 
const ll MOD=998244353;
const ll INF=1e18;
const int MAX=510000;
const double pi=acos(-1);
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};


int main(){
  cin.tie(0);
  ios::sync_with_stdio(false); 
  
  int a,b,c;
  cin >> a >> b >> c;
  int d=max({a,b,c});
  int f=min({a,b,c});
  int e=a+b+c-d-f;
  int ans=0;
  ans+=d-e;
  f+=ans;
  ans+=(d-f)/2;
  if(f%2!=d%2)ans+=2;
  
  cout << ans << endl;
  
  return 0;     
}