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.*;

public class Main{
  public static void main(String[] args){
    Scanner sc=new Scanner(System.in);
    long X=sc.nextLong();
    long K=sc.nextLong();
    long D=sc.nextLong();
    long div=0;
    long num=0;
    long ndiv=0;
    if(X>0){
    div=(long)Math.ceil
    ((double)X/(double)D);
    ndiv=Math.min(div,K);
    num=X-D*ndiv;
    }
    else if(X<0)
    {
      div=(long)Math.ceil
    ((double)Math.abs(X)/(double)D);
    ndiv=Math.min(div,K);
    num=X+D*ndiv;
    }
    K-=ndiv;
    long ans=0;
    if(K%2==0)
    {
      ans=Math.abs(num);
    }
    else
    {
      if(num<0)
      ans=Math.abs(num+D);
      else
      ans=Math.abs(num-D);
    }
    System.out.println(ans);
  }
}