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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Main {

  static long x, k, d, ans;

  public static void main(String[] args) throws Exception {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);

    String[] in = br.readLine().split(" ");
    x = Long.parseLong(in[0]);
    k = Long.parseLong(in[1]);
    d = Long.parseLong(in[2]);

    br.close();

    if (in[1].length() + in[2].length() > 17) {
      long p = x % d;
      boolean is_even = ((x / d) % 2 == 0);
      if ((k - x / d) % 2 == 0) {
        ans = p;
      } else {
        if (p < 0) {
          ans = p + d;
        } else {
          ans = p - d;
        }
      }
    } else if (Math.abs(x) <= k * d) {
      long p = x % d;
      boolean is_even = ((x / d) % 2 == 0);
      if ((k - x / d) % 2 == 0) {
        ans = p;
      } else {
        if (p < 0) {
          ans = p + d;
        } else {
          ans = p - d;
        }
      }
    } else {
      ans = Math.abs(x) - k * d;
    }

    out.println(Math.abs(ans));

    out.flush();
  }
}