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>

int min(int a,int b){
	if(a<b){return a;}
	else{return b;}
}

int max(int a,int b){
	if(a>b){return a;}
	else{return b;}
}

int main(){

	int A1,A2,A3;

	scanf("%d %d %d",&A1,&A2,&A3);

	int sum[6];

	sum[0]=max(A2-A1,A1-A2)+max(A3-A2,A2-A3);
	sum[1]=max(A3-A1,A1-A3)+max(A3-A2,A2-A3);
	sum[2]=max(A2-A3,A3-A2)+max(A3-A1,A1-A3);
	sum[3]=max(A2-A1,A1-A2)+max(A3-A1,A1-A3);
	sum[4]=max(A3-A1,A1-A3)+max(A1-A2,A2-A1);
	sum[5]=max(A3-A2,A2-A3)+max(A1-A2,A2-A1);

	int i;

	int ans=114514;

	for(i=0;i<6;i++){
	ans=min(ans,sum[i]);
	}

	printf("%d\n",ans);

	return 0;
}