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.io.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
class Main {
  public static void main(String[] args) {
    try {
        var stdin = new BufferedReader(new InputStreamReader(System.in));
        String line = null;
        while((line = stdin.readLine()) == null);
        var parsed = line.split(" ");
        Long X = Long.valueOf(parsed[0]);
        Long K = Long.valueOf(parsed[1]);
        Long D = Long.valueOf(parsed[2]);
        
        var sho = Math.abs(X) / D; // まず取りきる回数

        
        var joyo = Math.abs(X) % D; // 半端数
        var sign = X >= 0 ? 1 : -1; //初期場所の符号
        var maybemin = sign * joyo; // 最小値候補。半端数は符号を無視したので符号を戻してやる
        
                // とりきれなければ分かりやすい
        if (sho > K) {
            System.out.println(X - sign * D);
            return;
        }
        
        if (K - sho %2 == 0) { // 残り回数が偶数ならそのまま出す
            System.out.println(maybemin);
        } else {
            // 奇数なら最小値候補の符号と逆に動かす。最小値候補
            System.out.println(maybemin + sign * D);
        }
    } catch(Exception e) {
    }
   
  }
}