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.BigDecimal;
import java.util.*;

public class Main
{
    public static void main(String[] args)
    {
        //long start = System.currentTimeMillis();
        Scanner sc = new Scanner(System.in);
        long x = sc.nextLong();
        long k = sc.nextLong();
        long d = sc.nextLong();
        if (x == d)
        {
            if (k == 1) System.out.println(0);
            else System.out.println(x);
            System.exit(0);
        }
//        if (x >= k*d) System.out.println(x-(k*d));
//        else
//        {
//            long tmp = x/d;
//            long tmp1 = Math.abs(x-(d*tmp));
//            long tmp2 = Math.abs(x-(d*(tmp+1)));
//            if (tmp+1 >= k) System.out.println(tmp2);
//            else System.out.println(Math.min(tmp1,tmp2));
//        }

        BigDecimal bx = new BigDecimal(x);
        BigDecimal bk = new BigDecimal(k);
        BigDecimal bd = new BigDecimal(d);

        if (bx.compareTo(bk.multiply(bd)) >= 0) System.out.println(bx.subtract(bd.multiply(bk)));
        else
        {
            long tmp = x/d;
            BigDecimal tmp1 = bx.subtract(bd.multiply(new BigDecimal(tmp))).abs();
            BigDecimal tmp2 = bx.subtract(bd.multiply(new BigDecimal(tmp+1))).abs();
            if (tmp+1 >= k) System.out.println(tmp2);
            else
            {
                if (tmp1.compareTo(tmp2) >= 0) System.out.println(tmp2);
                else System.out.println(tmp1);
            }
        }
        sc.close();
//        long end = System.currentTimeMillis();
//        System.err.println((end-start)+"ms");
    }
}