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>
#include <boost/multiprecision/cpp_int.hpp>

using namespace std;
using namespace boost::multiprecision;

inline bool isEven(const int128_t a) { return 0 == a % 2; }

int main() {
	int128_t X, K, D;
	cin >> X >> K >> D;
	int128_t absX = abs(X);

	int128_t minAbs;


	if(absX >= D) {
		if(K <= absX / D) {
			minAbs = absX - K * D;
		} else {
			if(isEven(K)) {
				if(isEven(absX / D)) {
					minAbs = absX - (absX / D) * D;
					int128_t _abs = abs(minAbs - 2 * D);
					minAbs = min(minAbs, _abs);
				} else {
					minAbs = absX - (absX / D - 1) * D;
					int128_t _abs = abs(minAbs - 2 * D);
					minAbs = min(minAbs, _abs);
					_abs = abs(minAbs - 2 * D);
					minAbs = min(minAbs, _abs);
				}
			}
		}
	} else {
		if(isEven(K)) {
			minAbs = absX;
		} else {
			minAbs = abs(absX - D);
		}
	}

	cout << minAbs << endl;

	return 0;
}