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();

        long ans = 0;
        long kaisu = X / D;
        if (kaisu > K) {
            ans = Math.min(Math.abs(X - (D * K)), Math.abs(X + (D * K)));
        } else {
            if ((K-kaisu) %2 == 0) {
                ans = Math.min(Math.abs(X - (D * kaisu)), Math.abs(X + (D * kaisu)));
            } else {
                ans = Math.min(Math.abs(X - (D * (kaisu +1))), Math.abs(X + (D *(kaisu+1))));
            }
        }

        System.out.println(ans);
    }
}