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 <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <iomanip>
#include <string>
#include <bitset>
#include <cassert>
using namespace std;

int main()
{
	long long x, k, d;
	cin >> x >> k >> d;
	long long jumps = 0;
	if (abs(x) > d) {
		if (x >= 0) {
			jumps = x / d;
			if (k >= jumps) {
				k -= jumps;
				x = x % d;
			}
			else {
			
				cout << abs(x - d * k);
				return 0;
			}

		}
		else {
		
			jumps = abs(x) / d;
			if (k >= jumps) {
				k -= jumps;
				x = (-1) * (abs(x) % d);
			}
			else {
				
				cout << abs(x + d * k);
				return 0;
			}
		}
	}
	assert(abs(x) <= d);
	if (k%2==0) cout << abs(x);
	else {
		if (x < 0) cout <<  abs(x + d);
		else cout <<  abs(x - d);
	}
}