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 ll long long
#define ld long double
#define pii pair<int, int>
#define pll pair<ll, ll>
#define F first
#define S second
#define pb push_back
#define pp pop_back
#define ubo upper_bound
#define lbo lower_bound
#define VI vector<int>
#define VL vector<ll>
#define VP vector<pii >
#define all(x) x.begin(), x.end()
#define M vector<VI >

using namespace std;

int main(){

	ll x, k, d;
	cin >> x >> k >> d;

	x = abs(x);

	ll h = 0;
	for(int i=0; i<55; i++){
		ll t = 1;
		while((h+2*t)*d <= x) t*=2;
		if((h+t)*d <= x) h += t;
	}

	ll ans = x-min(h, k)*d;
	if(h < k){
		k -= h;
		if(k&1) ans = abs(ans-d);
	}

	cout << ans << '\n';

	return 0;
}