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 C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int main()
{
    long long X,K,D;
    scanf("%ld %ld %ld",&X,&K,&D);
    // 文字列の入力

	//reigai
	if (X == 0) {
		if (K % 2 == 0) {
			printf("%ld", labs(0));
			return 0;
		} else {
			printf("%ld", labs(D));
			return 0;
		}
	}

	if (X < 0) {
		while (X < 0) {
			X = X + D;
			K--;
			if (K == 0) {
				printf("%ld", labs(X));
				return 0;
			}
		}
		if (K % 2 == 0) {
			printf("%ld", labs(X));
			return 0;
		} else {
			printf("%ld", labs(X - D));
			return 0;
		}
	} else if (X > 0) {
		while (X > 0) {
			X = X - D;
			K--;
			if (K == 0) {
				printf("%ld", labs(X));
				return 0;
			}
		}
		if (K % 2 == 0) {
			printf("%ld", labs(X));
			return 0;
		} else {
			printf("%ld", labs(X + D));
			return 0;
		}
	}
}