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: Accepted)

import java.util.Scanner;

// 提出時にclass Main に書き換え
public class Main {
	public static void main(String[] args)  {
		// 標準入力を受け取る
		Scanner sc = new Scanner(System.in);
		long X = sc.nextLong(); // Long Xを受け取る
		long K = sc.nextLong(); // Long Kを受け取る
		long D = sc.nextLong(); // Long Dを受け取る

		sc.close(); // 標準入力を終了

		long x = X; // 現在の座標

		if(Math.abs(K * D) < Math.min(Math.abs(X + D / 2), Math.abs(X - D / 2))) {
			// X,Kに比べて、Dが極端に小さい場合
			x = Math.min(Math.abs(x + K * D), Math.abs(x - K * D));
			System.out.println(x);
			return; // プログラムを終了
		}

		for (int i = 0; i < K; i++) {
			x = Math.min(Math.abs(x + D), Math.abs(x - D));

			if(Math.abs(x) <= D / 2) {
				// 座標の絶対値の最小値

				// 残り回数が偶数か奇数か判定
				if((K - i - 1) % 2 == 0) {
					break;
				} else {
					// 奇数の場合は1回だけ動かす
					x = Math.min(Math.abs(x + D), Math.abs(x - D));
					break;
				}
			}
		}
		System.out.println(x);
	}

}