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 PI 3.1415926535897932
#define MOD 1000000007
#define INF 1e9

#define REP(i, n) for(int i = 0; i < (n); i++)
#define REP1(i, n) for(int i = 1; i <= (n); i++)
#define FOR(i, start, end) for(int i = (start); i < (end); i++)
#define ALL(v) (v).begin(), (v).end()

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef vector<bool> vb;


void solve()
{
    vi s(3); cin >> s[0] >> s[1] >> s[2];
    sort(ALL(s));
    int ans = 0;
    while(s[1] != s[2])
    {
        ans++;
        s[0]++;
        s[1]++;
    }
    while(s[0] < s[1])
    {
        ans++;
        s[0] += 2;
    }
    if(s[0] != s[1]) ans++;
    cout << ans << endl;
}

int main()
{
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    solve();

    return 0;
}