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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
using namespace std;

#include <math.h>
#include <iomanip>
#include <vector>
#include <queue>
#include <climits>
using Graph = vector<vector<int>>;
const double PI = 3.14159265358979323846;

int needed_kaisu(long long a, long long N, long long *A) {
	int count = 0;
	for (long long i = 0; i < N; i++) {
		count += (A[i]+a-1) / a -1;
	}
	return count;
}


int main() {


	string S;
	int N;
	long long X,K,D;

	cin >> X>>K>>D;

	long long minm, minp;

	X = abs(X);
	
	minp = X%D;
	minm = D - X%D;

	long long n = X / D;

	long long ans;

	if ((K - n) - ((K - n) / 2) * 2 == 0) {
		if (K >= n) {
			ans = minp;
		}
		else {
			ans = min(abs(X - K*D), abs(X + K*D));
		}
	}

	if ((K - n) - ((K - n) / 2) * 2 == 1) {
		if (K >= n+1) {
			ans = minm;
		}
		else {
			ans = min(abs(X - K*D), abs(X + K*D));
		}
	}

	cout << ans;

	cin >> N;

	return 0;


}