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.math.BigInteger;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);
		BigInteger x = new BigInteger(sc.next());
		BigInteger k = new BigInteger(sc.next());
		BigInteger d = new BigInteger(sc.next());
		BigInteger abs = x.abs();
		BigInteger goal = null;
		
		BigInteger div = x.divide(d);
		if (div.compareTo(k) > 0) {
			goal = abs.subtract(d.multiply(k));
			System.out.println(goal);
		} else if (div.compareTo(k) == 0) {
			goal = abs.subtract(d.multiply(k));
			System.out.println(goal);
		} else {
			// 距離縮める
			abs = abs.subtract(d.multiply(div));
			// 残り試行が奇数か偶数か
			BigInteger exa = k.subtract(div).mod(BigInteger.TWO);
			if (exa.equals(BigInteger.ZERO)) {
				// 偶数回
				goal = abs;
			} else {
				goal = abs.subtract(d).abs();
			}
			System.out.println(goal);
		}
	}
}