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 static java.lang.Math.*;

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

class Main {
    //無駄な計算を抑えるやつ
    static int mopi(List<Long> list, long X, long K, long D) {
        if(X > 0 && X / D >= K) {
            list.add(X - D * K);
            return -1;
        }
        return 0;
    }

    static int walkingTakashi(List<Long> list, long X, long K, long D) {
        list.add(X);
        if(X < 0) X = abs(X);
        if(mopi(list, X, K, D) != 0) return 0;
        if (X != 0) {
            long x = X;
            for (long i = K; i > 0; i--) {
                try {
                    x -= D;
                    list.add(abs(x));

                    if (x == 0)
                        break;
                    //0から遠ざかっているとき
                    int MAX = list.size();
                    if (abs(x) > abs(list.get(MAX - 2)))
                        break;
                    //xが異符号になったとき
                    if (signum(x) != signum(list.get(MAX - 2)))
                        break;

                } catch (ArithmeticException e) {
                    // TODO: handle exception
                    break;
                }
            }
        }
        return 0;
    }

    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        long X = std.nextLong();
        long K = std.nextLong();
        long D = std.nextLong();

        List<Long> list = new ArrayList<Long>();
        walkingTakashi(list, X, K, D);

        Collections.sort(list);
        System.out.println(list.get(0));
    }

}