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)

/*Infor {{{
 ******************************************************
 * > File Name: a.cpp
 * > Author: samcompu
 * > Mail: admin@samcompu.com
 * > Created Time: 2018/7/21 12:00:58
 ******************************************************
}}}*/

#include <bits/stdc++.h>

using namespace std;

/*Abbreviation {{{ */
#define FOR(i,a,b) for(int i = (int)(a); i <= (int)(b); i++)
#define ROF(i,a,b) for(int i = (int)(a); i >= (int)(b); i--)
#define cini(a) scanf("%d", &(a))
#define cinc(a) scanf("%c", &(a))
#define cinl(a) scanf("%lld", &(a))
#define cins(a) scanf("%s", (a))
#define cind(a) scanf("%lf", &(a))
#define coui(a) printf("%d", (a))
#define couc(a) printf("%c", (a))
#define coul(a) printf("%lld", (a))
#define cous(a) printf("%s", (a))
#define coud(a) printf("%f", (a))
#define SPA printf(" ")
#define LIN printf("\n")
#define ull unsigned long long int
/*}}}*/

/*Constant {{{*/
/*}}}*/

/*Class {{{*/
/*}}}*/

/*Variable {{{*/
int ans[10];
/*}}}*/

/*Declare {{{*/
/*}}}*/

/*Main {{{*/
int main() {
  int a, b, c;
  cini(a), cini(b), cini(c);
  ans[1] = abs(a - b) + abs(b - c);
  ans[2] = abs(a - c) + abs(c - b);
  ans[3] = abs(b - a) + abs(a - c);
  ans[4] = abs(b - c) + abs(c - a);
  ans[5] = abs(c - a) + abs(a - b);
  ans[6] = abs(c - b) + abs(b - a);
  int minCost = 0x7fffffff;
  FOR(i, 1, 6) minCost = min(minCost, ans[i]);
  coui(minCost), LIN;
  return 0;
}
/*}}}*/

/*Define {{{*/
/*}}}*/