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 <bits/stdc++.h>

#include <algorithm>
#include <array>
#include <chrono>
#include <cstdio>
#include <iostream>
#include <numeric>
#include <vector>

using namespace std;

#define ll long long
#define ld long double
#define P pair<ll, ll>
#define FOR(i, n, m) for (ll i = n; i < (ll)m; i++)
#define FORr(i, m, n) for (ll i = n; i >= (ll)m; i--)
#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)
#define sortAl(v) sort(v.begin(), v.end())
#define sortAlr(v)          \
  sort(v.begin(), v.end()); \
  reverse(v.begin(), v.end())
#define cout(n) cout << fixed << setprecision(n)

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  vector<ll> A(3);
  cin >> A[0] >> A[1] >> A[2];

  sort(A.begin(), A.end());

  ll sum = 0;
  for (int i = 1; i < A.size(); i++) {
    sum += A[i] - A[i - 1];
  }

  cout << sum << endl;

  return 0;
}