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: Wrong Answer)

#include<cstdio>
#include<algorithm>
using namespace std;
int main(void)
{
	long long x,k,d;
	long long l,s,c,a,b,t;
	scanf("%lld %lld %lld",&x,&k,&d);
	l=0; s=0; c=0; a=0; b=0;t=0;
	if(x<0){
		x=x*-1;
	}
	l=x-((x/d)*d);
	s=(x-((x/d+1)*d))*-1;
	if(s<l){
		c=x/d+1;
		b=1;
	}
	else{
		c=x/d;
		b=2;
	}
	a=k-c;
	if(c>k){
		printf("%lld\n",x-(k*d));
	}
	else{
		if(a%2==0){
			printf("%lld\n",l);
		}
		else{
			if(b==2){
				printf("%lld\n",s);
			}
			else{
				printf("%lld\n",l);
			}
		}
	}
	return 0;
}