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.*;
import java.util.function.*;
import java.math.*;
public class Main {
    public static void main(String[] args){
		Scanner s = new Scanner(System.in);
		long X = Long.valueOf(s.next());
		long K = Long.valueOf(s.next());
		long D = Long.valueOf(s.next());

		if(X < 0) {
			X = X * -1;
		}

		if(X/D >= K) {
			System.out.println(X - K * D);
			return;
		}

		long d = X / D;

		// 奇数
		if(d % 2 == 1) {
			// 奇数
			if(K % 2 == 1) {
				System.out.println(X % D);
				return;
			// 偶数
			}else {
				System.out.println(D - (X % D));
				return;
			}
		}else {
			// 奇数
			if(K % 2 == 1) {
				System.out.println(D - (X % D));
				return;
				// 偶数
			}else {
				System.out.println(X % D);
				return;
			}
		}
	}
}