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: Accepted)

#include <stdio.h>
#include <stdlib.h>
int d_cnt;
long int X, K, D, a, b, c;
int main(void) {
 
	scanf("%ld %ld %ld", &X, &K, &D);
	if(X < 0){
		d_cnt = 1;
		X = labs(X);
	}
	a = X % D;
	if(a < 0){
		a = a + D;
	}
	b = (X - a) / D;
	if(K > b){
	X = X - D * b;
 
	if((K-b) >= 2){
			if(((K - b) % 2) == 0){
				printf("%ld", X);
			}else{
				printf("%ld", labs(X-D));
			}
		}else if((K-b) == 1){
			printf("%ld", labs(X-D));
		}else if((K-b) == 0){
			printf("%ld", X);
		}
	}else{
		printf("%ld", (X-K*D));
	}
	return 0;
}