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<stdlib.h>
int a[3];
int num[6];
//void judge(int *a)
//{
//	int t;
//	if (a[0] < a[1]) { t = a[0]; a[0] = a[1]; a[1] = t; }
//	if (a[1] < a[2]) { t = a[1]; a[1] = a[2]; a[2] = t; }
//	if (a[0] < a[1]) { t = a[0]; a[0] = a[1]; a[1] = t; }
//}
int main(void)
{
	int sum = 0, min;
	for (int i = 0; i < 3; i++)
	{
		scanf("%d",&a[i]);
	}
	//judge(a);
	min = 1000;
	num[0] = abs(a[1] - a[0]) + abs(a[2] - a[1]);
	num[1] = abs(a[2] - a[0]) + abs(a[1] - a[2]);
	num[2] = abs(a[0] - a[1]) + abs(a[2] - a[0]);
	num[3] = abs(a[2] - a[1]) + abs(a[0] - a[2]);
	num[4] = abs(a[0] - a[2]) + abs(a[1] - a[0]);
	num[5] = abs(a[1] - a[2]) + abs(a[0] - a[1]);
	for (int i = 0; i < 6; i++)
	{
		if (min > num[i]) min = num[i];
	}
	printf("%d", min);
}