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;
typedef long long ll;
typedef unsigned long long ull;
#define pii pair<int,int>
#define pb push_back
#define fi first
#define se second
#define UP(a,b,c) for(ll (a)=(b);(a)<(c);(a)++)
#define UU(a,b,c) for(ll (a)=(b);(a)<=(c);(a)++)
#define DN(a,b,c) for(ll (a)=(b);(a)>(c);(a)--)
#define DU(a,b,c) for(ll (a)=(b);(a)>=(c);(a)--)

ll x,n,d;

void reset()
{
}

void input()
{
	cin>>x>>n>>d;
	
	x = abs(x);
	ll temp = x / d;
	
	temp = min(temp, n);
	n -= temp;
	x -= d*temp;
	
	n%=2;
	ll ans = abs(x);
	x-=d * n;
	
	cout<<abs(x)<<endl;
}

void solve()
{
}

void LetsRock()
{
	solve();	
}

int main()
{
	ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

	input();
	reset();
	LetsRock();
	
	return 0;
}