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.Comparator;
import java.util.Scanner;

public class Main {
    static int[] w = new int[100010];
    static int[] v = new int[100010];
    static int[][] dp = new int[110][100010];
    static int k;

    static int chmin(int a, int b) {
        if (a > b) {
            a = b;
            return a;
        }
        return a;
    }

    static int chmax(int a, int b) {
        if (a < b) {
            a = b;
            return a;
        }
        return a;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int a1 = sc.nextInt();
        int a2 = sc.nextInt();
        int a3 = sc.nextInt();

        int a1a2 = Math.abs(a1 - a2);
        int a1a3 = Math.abs(a1 - a3);
        int a2a3 = Math.abs(a2 - a3);

        if (a1a2 >= a1a3 && a1a2 >= a2a3)
            System.out.println(a1a3 + a2a3);
        else if (a1a3 >= a2a3 && a1a3 >= a1a2) {
            System.out.println(a2a3 + a1a2);
        } else
            System.out.println(a1a2 + a1a3);
    }
}