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 final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long x = Long.parseLong(sc.next());
        long k = Long.parseLong(sc.next());
        long d = Long.parseLong(sc.next());
        if (x > 0 && (x * 1.0 / k >= d)) {
            System.out.println(x - k * d);
            return;
        } else if (x < 0 && x * (-1.0) / k >= d) {
            System.out.println(x + k * d);
            return;
        }
        if (x > 0) {
            long num = (long)Math.ceil(x * 1.0 / d);
            long xx = x - ((num - 1) * d);
            if ((k - (num - 1)) % 2 == 0) {
                System.out.println(xx);
            } else {
                System.out.println(Math.abs(xx - d));
            }
        } else if (x < 0) {
            long num = (long)Math.ceil(x * (-1.0) / d);
            long xx = x + ((num - 1) * d);
            if ((k - (num - 1)) % 2 == 0) {
                System.out.println(Math.abs(xx));
            } else {
                System.out.println(xx + d);
            }
        } else {
            if (k % 2 == 0) {
                System.out.println(0);
            } else {
                System.out.println(d);
            }
        }
    }
}