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>

using namespace std;

int main() {
  
  long x,k,d;
  cin>>x>>k>>d;
  if(x>=0)
  {
      long n = x/d+1;
      if(n<=k)
      {
          
      long long int a1 = x -(n-1)*1LL*(d);
      long long int a2 = x-(n)*1LL*d;
      long long int m = min(abs(a1),abs(a2));
      cout<<m;
      }
      else
      {
          cout<<(x-k*1LL*d);
      }
  }
  else
  {
      long long int n= abs(x/d+1);
      if(n<=k)
      {
           long long int a1 = x +(n-1)*1LL*(d);
      long long int a2 = x+(n)*1LL*d;
      long long int m = min(abs(a1),abs(a2));
      cout<<m;
      }
      else
      cout<<(x+k*1LL*d);
  }
   
}