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<bits/stdc++.h>
using namespace std;
int main()
{
	long long x,k,d;
	cin>>x>>k>>d;
	long long y=abs(x);
	long long j=y/d;
	if(x==0)
	{
		if(k%2)
		cout<<d;
		else
		cout<<0;
	}
	else if(x<0)
	{
		if(j>=k)
		{
			x+=(d*k);
			cout<<abs(x);
		}
		else
		{
			x+=(d*j);
			k=k-j;
			if(k%2==0)
			{
				cout<<abs(x);
			}
			else
			cout<<abs(x+d);
		}
	}
	else if(x>0)
	{
		if(j>=k)
		{
			x-=(d*k);
			cout<<abs(x);
		}
		else
		{
			x-=(d*j);
			k=k-j;
			if(k%2==0)
			{
				cout<<abs(x);
			}
			else
			cout<<abs(x-d);
		}
	}
	return 0;
}