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 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
	    Scanner sc = new Scanner(System.in);
	    long x = sc.nextLong();
	    long k = sc.nextLong();
	    long d = sc.nextLong();
	    List wk = new ArrayList();
	    if (x == d) {
	    	if (k % 2 == 0)  System.out.println(x);
	    	if (k % 2 == 1)  System.out.println(0);
	    } else if (Math.abs(k * d) < Math.abs(x)) {
	    	if (x < 0) System.out.println(Math.abs(x + k*d));
	    	if (x >= 0) System.out.println(Math.abs(x - k*d));
	    } else {
	    	boolean flag = false;
		    for (long i = 0; i < k; i++) {
		    	if (Math.abs(x - d) > Math.abs(x + d)) {
		    		x = x + d;
		    	} else {
		    		x = x - d;
		    	}
		    	if (flag) {
		    		break;
		    	}
		    	// 同じ場所の往復
		    	if (wk.contains(x)) {
		    		if ((k - i - 1) % 2 == 0) {
		    			break;
		    		} else {
		    			flag = true;
		    		}
		    	}
		    	wk.add(x);
		    }
		    System.out.println(Math.abs(x));
	    }
	}
}