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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static String Y = "Yes";
	public static String N = "No";
	public static long MOD = (long) (Math.pow(10, 9) + 7);
	public static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) {
		int a = ni();
		int b = ni();
		int c = ni();

		int[] abc = { a, b, c };

		Arrays.sort(abc);

		int count = 0;
		count += abc[1] - abc[0] + abc[2] - abc[1];
		out(count);

	}

	/*
	 * 以下メソッド集
	 */

	static int ketasuu(int n) {
		String str = "" + n;
		return str.length();
	}

	static int abcount(String str) {
		int count = 0;
		int len = str.length();
		for (int i = 0; i < len - 1; i++) {
			if ("AB".equals(str.substring(i, i + 2))) {
				count++;
			}
		}
		return count;
	}

	static int ni() {
		return sc.nextInt();
	}

	static long nl() {
		return sc.nextLong();
	}

	static double nd() {
		return sc.nextDouble();
	}

	static String n() {
		return sc.next();
	}

	static char[] nc() {
		return sc.next().toCharArray();
	}

	static int kaijo(int n) {
		if (n == 0 || n == 1) {
			return 1;
		} else {
			return n * kaijo(n - 1);
		}
	}

	static int fib(int n) {
		return (n == 1 || n == 0) ? n : fib(n - 2) + fib(n - 1);
	}

	static long lcm(long m, long n) {

		return m * n / gcd(m, n);
	}

	static int lcm(int m, int n) {

		return m * n / gcd(m, n);
	}

	static long gcd(long a, long b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	static int gcd(int a, int b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	static void out(String info) {
		System.out.println(info);
	}

	static void out(int info) {
		System.out.println(info);
	}

	static void out(double info) {
		System.out.println(info);
	}

	static void out(long info) {
		System.out.println(info);
	}

	static void out(char info) {
		System.out.println(info);
	}

	static void out(Object obj) {
		System.out.println(obj.toString());
	}

	static void outn(String info) {
		System.out.print(info);
	}

	static void outn(int info) {
		System.out.print(info);
	}

	static void outn(double info) {
		System.out.print(info);
	}

	static void outn(long info) {
		System.out.print(info);
	}

	static void outn(char info) {
		System.out.print(info);
	}

	static double max(double d, double e) {
		return Math.max(d, e);
	}

	static long max(long a, long b) {
		return Math.max(a, b);
	}

	static double min(double d, double e) {
		return Math.min(d, e);
	}

	static long min(long a, long b) {
		return Math.min(a, b);
	}
}

class XY {
	int h;
	int c;

	XY(int h, int c) {
		this.h = h;
		this.c = c;
	}
}