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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
void main(void){
    int a[3];
    int i,j,tmp,max;
    int s[3];

    for(i=0;i<3;i++){
        if(scanf("%d",&a[i]) == 1);
    }

    for(i=0;i<3;i++){
        for(j=i+1;j<3;j++){
        if(a[i]<a[j]){
            tmp = a[i];
            a[i] = a[j];
            a[j] = tmp;
        }
        }
    }

    s[0] = a[0] - a[1];
    s[1] = a[1] - a[2];
    s[2] = a[0] - a[2];

    for(i=0;i<3;i++){
         for(j=i+1;j<3;j++){
        if(s[i]>s[j]){
            tmp = s[i];
            s[i] = s[j];
            s[j] = tmp;
        }
         }
    }

    for(i=0;i<3;i++){
        printf("%d  %d\n",a[i],s[i]);
    }

    max = s[0] + s[1];

    printf("%d\n",max);
}