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){
        Main main = new Main();
        main.input();
        main.solve();
        main.output();
    }

    long X_start;
    long K_times;
    long D_reach;
    long x;
    long time;
    long result;

    private void input() {
        Scanner sc = new Scanner(System.in);
        X_start = sc.nextLong();
        K_times = sc.nextLong();
        D_reach = sc.nextLong();

        x = X_start;
        time = 0;
    }

    private void solve() {
        approach();
        if (time == K_times) {
            return;
        }
        if ((K_times-time) % 2 == 0) {
            return;
        } else {
            if (x < 0) {
                x = x + D_reach;
            } else {
                x = x - D_reach;
            }
        }
    }

    private void approach() {
        long range = x < 0 ? -x : x;
        long t_approaching = range / D_reach;
        if (K_times < t_approaching) {
            time = K_times;
            if (x < 0) {
                x = x + D_reach*K_times;
            } else {
                x = x - D_reach*K_times;
            }
        } else {
            time = t_approaching;
            if (x < 0) {
                x = x + D_reach*t_approaching;
            } else {
                x = x - D_reach*t_approaching;
            }
        }
    }

    private void output() {
        if (x < 0) {
            System.out.println(-x);
        } else {
            System.out.println(x);
        }
    }
}