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<string.h>
#include<stdlib.h>
#include<math.h>
#define PI 3.141592653589793
#define MOD 1000000007
//#define DEBUG

//qsort用、昇順
int compare_int(const void *a, const void *b)
{
	return *(int*)a-*(int*)b;
}
//qsort(array, 10, sizeof(int), compare_int)



int main(){
	long long int x, k, d;
	scanf("%lld%lld%lld", &x, &k, &d);

	long long int syo=0, ans=x;
	if(x < 0){
		x = x*(-1);
	}
	syo = x/d;

	if(syo >= k){
		ans = x - d*k;
	}
	else{
		k = k - syo;
		x = x - d*syo;

		if(k%2 == 0){
			ans = x;
		}
		else{
			ans = d - x;
		}
	}

	printf("%lld\n", ans);
	
	return 0;
}