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>
#include <vector>
#include <iostream>
#include<algorithm>
#include<string>
#include <map>
#include <queue>
#include <stack>
#include<set>

#define DIV 1000000007
#define TE 2e5
using namespace std;
using ll = long long;
using ldb = long double;

int main() {
	ll X, K, D; cin >> X >> K >> D;
	ll ans = 0;
	X = abs(X);
	if ((abs(X)+D-1)/D <= K) {
		ll p = abs(X) / D;
		if (p % 2 == K % 2)
			ans = abs(p * D - abs(X));
		else
			ans = abs((p + 1) * D - abs(X));
		//ans = min(abs((abs(X) / D) * D - abs(X)), abs((abs(X) + D - 1) / D) * D - abs(X));
		//cout << (abs(X) / D) % 2 <<" "<< K % 2 << endl;
	}
	else ans = abs(abs(X) - K * D);
	cout << ans << endl;
}