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_base::sync_with_stdio(false); 
    cin.tie(NULL); 
    
      long long n,i,j,k,sum=0,count=0,ans=0,x=0,y=0,l,r,m;
      cin>>n>>m>>k;
      if(n==0)
      {
          if(m%2==0)
          cout<<0;
          else
          cout<<k;
          return 0;
      }
      if(n>0)
      {
          if(n<k)
          {ans=n; x=0;}
          else
          {ans=n-(n/k)*k; x=n/k;}
          
          if(x<m)
          {
               sum=abs(ans-k);
              m=m-x;
              if(m%2==1)
              cout<<sum;
              else
              cout<<ans;
               
          }
          else
          {
             cout<<n-k*m; 
          }
          
      }
      else
      {
          n=-n;
        if(n<k)
          {ans=n; x=0;}
          else
          {ans=n-(n/k)*k; x=n/k;}
          
          if(x<m)
          {
             sum=abs(ans-k);
              m=m-x;
              if(m%2==1)
              cout<<sum;
              else
              cout<<ans;   
          }
          else
          {
             cout<<n-k*m; 
          }
      }
	return 0;
}