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>

void insert(int *a,int i,int j){//i番めの要素をj番めの要素のとこに挿入(i,j is index)i>j
    int k,val;
    val = a[i];
    if(i==j){
        return;
    }
    for(k=i;k>j;k+=-1){
        a[k] = a[k-1];
    }
    a[j] = val;
    return;
}

int *insertion_sort(int *a,int size){
    int i,insert_place,val;
    for(i=0;i<size;i++){
        insert_place = i-1;
        while(insert_place>=0 && a[i]<a[insert_place]){
            insert_place += -1;
        }
        insert_place += 1;
        insert(a,i,insert_place);
    }
    return a;
}

int main(void){
    int a[3];
    scanf("%d %d %d",&a[0],&a[1],&a[2]);
    int *b;
    b= insertion_sort(a,3);
    int i,counter=0;
    for(i=0;i<2;i++){
        counter = counter + a[i]-a[i+1];
    }
    printf("%d\n",counter*-1);
    return 0;

}