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: Accepted)

#include <stdio.h>
#include <math.h>

int main(void)
{
    int a[3];
    int i,j,v, cost;
    int aa[3][3];
    for (i=0;i<3;i++) {
        scanf("%d", &a[i]);
    }

    for (i=1;i<3;i++) {
        v = a[i];
        j = i-1;
        while (j>=0&&a[j]>v){
            a[j+1]=a[j];
            j--;
        }
        a[j+1]=v;
    }

    if (abs(a[1]-a[0])>=abs(a[2]-a[1]) && abs(a[1]-a[0])>=abs(a[0]-a[2])){
        cost = abs(a[1]-a[0])+abs(a[2]-a[1]);
    }

    if (abs(a[2]-a[1])>=abs(a[1]-a[0]) && abs(a[2]-a[1])>=abs(a[0]-a[2])) {
        cost = abs(a[1]-a[2])+abs(a[0]-a[1]);      
    }

    if (abs(a[0]-a[2])>=abs(a[2]-a[1]) && abs(a[0]-a[2])>=abs(a[1]-a[0])) {
        if (abs(a[0]-a[1])<abs(a[2]-a[1])) {
             cost = abs(a[1]-a[2])+abs(a[0]-a[1]);       
        } else {
             cost = abs(a[1]-a[0])+abs(a[2]-a[1]);
        }

    }
    


    printf("%d\n", cost);
    return 0;
}