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)

#define _USE_MATH_DEFINES
#include <cassert>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <complex>
#include <cmath>
#include <numeric>
#include <bitset>
#include <functional>

using namespace std;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}

typedef long long int64;
typedef pair<int, int> ii;
const int INF = 1 << 29;
const int MOD = 1e9 + 7;

int main() {
  vector<int> p(3), a(3);
  for (int i = 0; i < 3; ++i) p[i] = i;
  for (int i = 0; i < 3; ++i) {
    scanf("%d", &a[i]);
  }
  int ret = INF;
  do {
    int cost = 0;
    for (int i = 1; i < 3; ++i) {
      cost += abs(a[p[i - 1]] - a[p[i]]);
    }
    ret = min(ret, cost);
  } while (next_permutation(p.begin(), p.end()));
  printf("%d\n", ret);
  return 0;
}