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 <cstring>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <numeric>
#include <utility>
#include <unordered_map>
#include <unordered_set>

using namespace std;

#ifndef debug_on
#define debug_on 0
#endif

#define dbg \
    if(debug_on) cerr << "DEBUG:"
#define ast(cond)           \
    do {                    \
        if((cond) == false) \
        {                   \
            if(debug_on)    \
            cerr << "ASSERT" << #cond " failed at line:" << __LINE__ << endl;   \
            exit(1);        \
        }                   \
    }while(0)

int V[3];

int main(int, char *[])
{
    cin >> V[0] >> V[1] >> V[2];
    sort(V, V+3);
    V[2] -= V[1];
    V[0] -= V[1];
    V[1] -= V[1];

    int count[2] = {0, 0};
    while(V[0] < 0)
    {
        if(V[0] <=  -2)
        {
            V[0] += 2;
            ++count[0];
        }
        else if(V[0] == -1)
        {
            ++V[0];
            count[0] += 2;
        }
    }
    count[1] = V[2];


    cout << (count[0] + count[1]) << endl;

    return 0;
}