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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;

/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x, long long int y) {
	return(x + y - 1) / y;

}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;

	if (number2 > number1) {
		m = number2;
		n = number1;
	}
	long long int s = -1;
	while (s != 0) {
		s = m % n;
		m = n;
		n = s;
	}
	return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;
	return m / lcm(m, n) * n;
}
int dp[120][10200] = {};
int main() {
	int d[3];
	for (int i = 0; i < 3; i++) {
		int a;
		cin >> a;
		d[i] = a;
	}
	int sum = 100000;
	for (int i = 0; i < 3; i++) {
		int k1 = 0;
		for (int j = 0; j < 3; j++) {
			if (i == j) {
				continue;
			}
			for (int k = 0; k < 3; k++) {
				if (i == k || j == k) {
					continue;
				}
				sum = min(abs(d[j]-d[i])+abs(d[k]-d[j]), sum);

			}
		}
	}
	cout << sum;
}