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 int long long
#define pii pair<int,int>
#define vi vector<int>
#define pb push_back
#define ff first
#define ss second
#define init(a, x) memset(a, x, sizeof(a))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

using namespace std;

signed main(){
	
	int x, k, d;

	cin >> x >> k >> d;
	x = abs(x);

	int quo = x / d;
	
	int res1 = 1e18, res2 = 1e18;
	if ((quo <= k) && ((k - quo) % 2 == 0) ){
		res1 = x - (quo * d);		
	} 
	if ((quo <= k + 1) && ((k - quo - 1) % 2 == 0)){
		res2 = x - ((quo + 1) * d);

	} 
	if (quo > k){
		res1 = x - (k * d);
	}

	int res = min(abs(res1), abs(res2));
	cout << res << endl;
	return 0;
}