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.util.*;
public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        String x_k_d [] = s.split(" ", -1);
        long x = Long.parseLong(x_k_d[0]);
        long k = Long.parseLong(x_k_d[1]);
        long d = Long.parseLong(x_k_d[2]);
        
        //移動した後の位置pos、残り移動回数a
        long pos = x;
        long a = k;
        long b = 0;
        long c = 0;

        //出力は絶対値なのでposを正にして計算する
        if(x < 0){
            pos *= -1;
        }
        //0に近づけるように移動させる
        if(x / d < k){
            a -= pos /d;
            pos = pos % d;
            
        }else{
            a = 0;
            pos -= d * k;
        }
        //System.out.println(pos);
        //System.out.println(a);
        
        if(a == 0){
            System.out.println(pos);
        }else{
            b = pos;
            c = (pos - d) * -1;
            if(a % 2 == 0){
                System.out.println(b);
            }else if(a % 2 == 1){
                System.out.println(c);
            }
        }
    }
}