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[]) {

        // 入力
        Scanner sc = new Scanner(System.in);
        long x = Math.abs(Long.parseLong(sc.next()));
        long k = Long.parseLong(sc.next());
        long d = Long.parseLong(sc.next());
        sc.close();

        // 主処理
        long result = 0;
        long base = d * 2;
        if ((x % d == 0 && k <= x / d) || (x % d != 0 && k <= (x / d + 1))) {
            result = Math.abs(x - k * d);
        } else if (k % 2 == 0) {
            long mlt = (long) Math.ceil((double) x / base);
            long right = base * mlt;
            long left = right - base;
            long plus = Math.min(Math.abs(x + right), Math.abs(x + left));
            long minus = Math.min(Math.abs(x - right), Math.abs(x - left));
            result = Math.min(plus, minus);
        } else if (k == 1) {
            long plus = Math.abs(x + d);
            long minus = Math.abs(x - d);
            result = Math.min(plus, minus);
        } else {
            long mlt = (long) Math.ceil(Math.ceil((double) x / d) / 2.0);
            long right = base * mlt + d;
            long left = right - base;
            long plus = Math.min(Math.abs(x + right), Math.abs(x + left));
            long minus = Math.min(Math.abs(x - right), Math.abs(x - left));
            result = Math.min(plus, minus);
        }

        // 出力
        System.out.println(result);
    }
}