Problem:
Score : 300 points

Problem Statement

Takahashi, who lives on the number line, is now at coordinate X. He will make exactly K moves of distance D in the positive or negative direction.

More specifically, in one move, he can go from coordinate x to x + D or x - D.

He wants to make K moves so that the absolute value of the coordinate of the destination will be the smallest possible.

Find the minimum possible absolute value of the coordinate of the destination.

Constraints

-10^{15} \leq X \leq 10^{15}

1 \leq K \leq 10^{15}

1 \leq D \leq 10^{15}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X K D

Output

Print the minimum possible absolute value of the coordinate of the destination.

Sample Input 1

6 2 4

Sample Output 1

2

Takahashi is now at coordinate 6. It is optimal to make the following moves:

Move from coordinate 6 to (6 - 4 =) 2.

Move from coordinate 2 to (2 - 4 =) -2.

Here, the absolute value of the coordinate of the destination is 2, and we cannot make it smaller.

Sample Input 2

7 4 3

Sample Output 2

1

Takahashi is now at coordinate 7. It is optimal to make, for example, the following moves:

Move from coordinate 7 to 4.

Move from coordinate 4 to 7.

Move from coordinate 7 to 4.

Move from coordinate 4 to 1.

Here, the absolute value of the coordinate of the destination is 1, and we cannot make it smaller.

Sample Input 3

10 1 2

Sample Output 3

8

Sample Input 4

1000000000000000 1000000000000000 1000000000000000

Sample Output 4

1000000000000000

The answer can be enormous.

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
public class Main extends PrintWriter {
    Main()
    {
        super(System.out);
    }
    public static void main(String args[]) {
      Main o=new Main();
      o.main();
      o.flush();
      o.close();
    }
    void main()
    {
        Scanner sc=new Scanner(System.in);
        long x=sc.nextLong();
        long k=sc.nextLong();
        long d=sc.nextLong();
        long rem=x%d;
        long quo=x/d;
        if(x<0)
        quo*=-1;
        if(k<quo)
        {
            if(x>0)
            {
                println(Math.abs(x-(k*d)));
            }
            else
            {
                println(Math.abs(x+(k*d)));
            }
        }
        else
        {
            if(x>=0)
            {
                long v1=rem-d;
                long rem1=k-quo;
                if(rem1%2==0)
                println(Math.abs(rem));
                else
                println(Math.abs(v1));
            }
            else
            {
                long v2=rem+d;
                long rem2=k-quo;
                if(rem2%2==0)
                println(Math.abs(rem));
                else
                println(Math.abs(v2));
            }
        }
        sc.close();
    }
}