Problem:
Score : 100 points

Problem Statement

You have three tasks, all of which need to be completed.

First, you can complete any one task at cost 0.

Then, just after completing the i-th task, you can complete the j-th task at cost |A_j - A_i|.

Here, |x| denotes the absolute value of x.

Find the minimum total cost required to complete all the task.

Constraints

All values in input are integers.

1 \leq A_1, A_2, A_3 \leq 100

Input

Input is given from Standard Input in the following format:

A_1 A_2 A_3

Output

Print the minimum total cost required to complete all the task.

Sample Input 1

1 6 3

Sample Output 1

5

When the tasks are completed in the following order, the total cost will be 5, which is the minimum:

Complete the first task at cost 0.

Complete the third task at cost 2.

Complete the second task at cost 3.

Sample Input 2

11 5 5

Sample Output 2

6

Sample Input 3

100 100 100

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <utility>
#include <iomanip>

#define ll long long int
#define pb push_back
#define mk make_pair
#define pq priority_queue

using namespace std;
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
const int inf = 1e9;
const ll linf = 1LL << 50;

int main(int argc, char const* argv[])
{
	int a1, a2, a3;
	cin >> a1 >> a2 >> a3;
	int maxi = max(a1, max(a2, a3));
	int mini = min(a1, min(a2, a3));
	int mid = 0;
	if(a1 != maxi && a1 != mini)mid = a1;
	if(a2 != maxi && a2 != mini)mid = a2;
	if(a3 != maxi && a3 != mini)mid = a3;
	cout << maxi - mini << endl;
	return 0;
}