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  = Math.abs(sc.nextLong());
		long k  = sc.nextLong();
		long d  = sc.nextLong();
		long ans = 0;

		//0に近づけるだけで回数分動く場合
		if(k * d <= x) {
			System.out.println(x - (k * d));
			sc.close();
			return;
		}
		
		//最初にできるだけ0に近づける
		long devide = x / d;
		long rest = x - (d * (x / d));
		
		//k - divideが偶数なら何もしない
		if((k - devide) % 2 == 0) {
			ans = rest;
		}
		
		//k - divideが奇数なら1度だけ正か負に動かして絶対値の小さいほう
		if((k - devide) % 2 != 0) {
			long culcA = rest + d;
			long culcB = rest - d;			
			ans = Math.min(Math.abs(culcA),Math.abs(culcB));
		}
		
		System.out.println(ans);
		sc.close();
	}
}