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<bits/stdc++.h>
using namespace std;
#define ll long long
#define forn(i,a,b) for(ll i=a;i<b;i++)
#define pb push_back
#define mp make_pair

int main()
{
	ll x,k,d; cin >> x >> k >> d; ll ans;

	if(x > 0)
	{
		if((d*k) > x )
		{
			ll var = x%d;
			ll neg = var - d;
			ll a = x/d;
			k -= a;
			// cout << k << endl;
			if(k%2 == 0)
				ans = var;
			else
				ans = -(neg);
		}
		else
			ans = x - (d*k);
	}

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

	if(x < 0)
	{
		x = -(x);

		if((d*k) > x )
		{
			ll var = x%d;
			ll neg = var + d;
			ll a = x/d;
			k -= a;
			// cout << k << endl;
			if(k%2 == 0)
				ans = (neg);
			else
				ans = var;
		}
		else
			ans = (-x) + (d*k);
	}

	cout << ans << endl;
}