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: Runtime Error)

import java.io.IOException;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException{
		Scanner sc = new Scanner(System.in);
		Integer X = sc.nextInt();
		Integer K = sc.nextInt();
		Integer D = sc.nextInt();
		sc.close();

		Integer temp[] = new Integer[K];

		for(int i=0; i<K; i++) {
			if(X>0){
				X = X - D;
				temp[i] = X;
			}
			else if(X<0) {
				X = X + D;
				temp[i] = X;
			}
		}

		int min = 0;
		for(int i=0; i<K; i++) {
			if(i>=1) {
				if(Math.abs(temp[i])>Math.abs(temp[i-1])) {
					min = i-1;
				}
			}	
		}

		int result = 0;
		if((K - min + 1)%2==0) {
			//偶数回
			result = Math.abs(temp[min]);
		}
		else {
			//奇数回
			int l = temp.length;
			if(min - 1 < 0) {
				result = Math.abs(temp[min]);
			}
			else if(min + 1 > l -1) {
				result = Math.abs(temp[min]);
			}
			else {
			result = ( Math.abs(temp[min-1]) < Math.abs(temp[min+1]) ) ?
					Math.abs(temp[min-1]) : Math.abs(temp[min+1]);
			}
			}

		System.out.println(result);
	}
}