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>
#define pb push_back
#define INF 1<<30
#define all(x) x.begin(),x.end()
#define ms(a,v) memset(a,v,sizeof a)
#define II ({int a; scanf("%d", &a); a;})
#define LL  ({ll  a; scanf("%lld", &a); a;})
#define EPS 1e-10
#define pi acos(0.0)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef vector<int> vi ;
typedef vector<ll>vl;

#define FOR(i,n) for( __typeof(n)i = 0 ; i < n ; i++)
#define FORL(i,a, b) for( __typeof(n)i = a ; i <= b ; i++)
#define FORR(i,a, b) for( __typeof(n)i = a ; i >= b ; i--)
#define For(i,n) for( __typeof(n)i = 1 ; i <= n ; i++)
#define forstl(i,n) for(__typeof(n.begin())i = n.begin();i!=n.end();i++)
#define forab(i,a,b) for( __typeof(b) i = a ; i <= (b) ; i++ )
#define forba(i,b,a) for( __typeof(b) i = b ; i >= (a) ; i-- )

int x[4] = {0 , 0 , -1, 1};
int y[4] = {1, -1, 0  , 0};



void solve()
{
	ll x, k , d;
	cin >> x >> k >> d;

	x = abs(x);

	ll t = x / d;
	if (t > k)
	{
		cout << (x - k * d) << endl;
	}
	else
	{
		k -= t;
		t = x - t * d;
		if (k % 2 == 0)
		{
			cout << t << endl;
		}
		else
		{
			cout << abs(t - d) << endl;
		}
	}
}

int main()
{


	int T = 1;
	//cin >> T;
	// cin.ignore(); must be there when using getline(cin, s)

	//freopen("input.txt", "r", stdin);

	// Printing the Output to output.txt file
	//freopen("output.txt", "w", stdout);

	while (T--)
	{
		solve();
	}
	return 0;
}