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.*;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        BigInteger TWO = new BigInteger("2");
        Scanner sc = new Scanner(System.in);
        BigInteger x = new BigInteger(sc.next());
        BigInteger k = new BigInteger(sc.next());
        BigInteger d = new BigInteger(sc.next());
        if (x.abs().compareTo(k.multiply(d)) >= 0) {
            if (x.compareTo(BigInteger.ZERO) >= 0) System.out.println(x.subtract(k.multiply(d)));
            else System.out.println(x.add(k.multiply(d)));
        // } else if (x.compareTo(BigInteger.ZERO) >= 0) {
        //     BigInteger[] dr = x.divideAndRemainder(d);
        //     if (k.remainder(TWO).equals(BigInteger.ZERO)) {
        //         if (dr[0].remainder(TWO).equals(BigInteger.ZERO)) System.out.println(dr[1]);
        //         else System.out.println(dr[1].subtract(d));
        //     }
        } else {
            BigInteger[] dr = x.abs().divideAndRemainder(d);
            if (k.remainder(TWO).equals(BigInteger.ZERO)) {
                if (dr[0].remainder(TWO).equals(BigInteger.ZERO)) System.out.println(dr[1].abs());
                else System.out.println(d.subtract(dr[1]).abs());
            }
        }
    }
}