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: Wrong Answer)

import java.util.Scanner;

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();

    System.out.println(solve(Math.abs(x), k, d));
  }
  static long solve(long x, long k, long d){
    long a = x / d;
    if(a == 0){
      if(Math.abs(x) <= Math.abs(x - d)){
        if(k % 2 == 0){
          return Math.abs(x);
        } else {
          return Math.abs(Math.abs(x) - d);
        }
      }
      x = x - d;
      k--;
      if(k % 2 == 0){
        return Math.abs(x);
      } else {
        return Math.abs(Math.abs(x) - d);
      }
    } else if(a > k){
      x = x - (k * d);
      return Math.abs(x);
    } else if(a == k){
      return 0;
    } else {
      // (a < k)
      x = x - (a * d);
      k -= a;
      if(k % 2 == 0){
        return Math.abs(x);
      } else {
        return Math.abs(Math.abs(x) - d);
      }
    }
  }
}