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>
using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long x, k, d;
  cin >> x >> k >> d;
  long long res;
  if (x < 0) {
    // x + d * k >= 0
    long long nk = -x / d;
    if (x + d * nk < 0)
      ++nk;
    if (nk > k)
      res = x + d * k;
    else {
      long long rem = nk - k;
      if (rem % 2 == 0)
        res = x + d * nk;
      else
        res = x + d * (nk - 1);
    }
  } else {
    // x - d * k <= 0
    long long nk = x / d;
    if (x - d * nk > 0)
      ++nk;
    if (nk > k)
      res = x - d * k;
    else {
      long long rem = nk - k;
      if (rem % 2 == 0)
        res = x - d * nk;
      else
        res = x - d * (nk - 1);
    }
  }
  cout << abs(res) << '\n';
  return 0;
}