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>
#define ff first
#define ss second
#define ll long long
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;

void solve() {

   ll x, k, d;
   cin >> x >> k >> d;
   x = abs(x);

   if (k >= x / d) {
      int r = x % d, m = k - x / d;
      if ((k - (x + d - 1) / d) % 2 == 0) {
         cout << min({x % d, x - x % d}) << '\n';
      } else {
         cout << min({x % d, x - x % d}) + d << '\n';
      }
   } else {
      cout << x - k * d << '\n';
   }

}

int main() {
#ifndef ONLINE_JUDGE
   freopen("inp.txt", "r", stdin);      freopen("outp.txt", "w", stdout);
#endif
   ios::sync_with_stdio(false);   cin.tie(NULL);   cout.tie(NULL);

   solve();

   return 0;
}