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;
typedef long long ll;
#define mod 1000000007
main()
{
     ll x,k,d,ans=0,s;
     cin>>x>>k>>d;
     ll m=x%d;
     if(m<0)m+=d;
     //cout<<m<<endl;
     s=abs(m-x)/d;
     ll m1=m-d;
     ll s1=abs(m1-x)/d;
     if(s1<s)
     {
         s=s1;
     }
     //cout<<s<<endl;
     if(x>0)
         {
             ans=x-s*d;
         }
         else
         {
             ans=x+s*d;
         }
     if(k<s)
     {
         if(x>0)
         {
             ans=x-k*d;
         }
         else
         {
             ans=x+k*d;
         }
         cout<<abs(ans)<<endl;
     }
     else if((k-s)%2==0)
     {

         cout<<abs(ans)<<endl;
     }
     else
     {
         if(abs(ans-d)<abs(ans+d))
         {
             cout<<abs(ans-d)<<endl;
         }
         else
         {
            cout<<abs(ans+d)<<endl;
         }
     }
}