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>
using namespace std;
using ll=long long;
int main(){
  ll x,k,d,p;
  cin>>x>>k>>d;
  if(log10(k)+log10(d)>16){
    ll a=min(abs(x%d),abs(abs(x%d)-d));
    if(a==abs(x%d)){
      p=abs((a-x)/d);
    }else{
      p=abs((-a-x)/d);
    }if(p%2==k%2){
      cout<<min(abs(x%d),abs(abs(x%d)-d))<<endl;
    }else{
      cout<<max(abs(x%d),abs(abs(x%d)-d))<<endl;
    }
  }else if(x<0){
    if(x+k*d<0){
      cout<<abs(x+k*d)<<endl;
    }else{
      ll a=min(abs(x%d),abs(abs(x%d)-d));
      if(a==abs(x%d)){
        p=abs((a-x)/d);
      }else{
        p=abs((-a-x)/d);
      }if(p%2==k%2){
        cout<<min(abs(x%d),abs(abs(x%d)-d))<<endl;
      }else{
        cout<<max(abs(x%d),abs(abs(x%d)-d))<<endl;
      }
    }
  }else{
    if(x-k*d>0){
      cout<<x-k*d<<endl;
    }else{
      ll a=min(abs(x%d),abs(abs(x%d)-d));
      if(a==abs(x%d)){
        p=abs((a-x)/d);
      }else{
        p=abs((-a-x)/d);
      }if(p%2==k%2){
        cout<<min(abs(x%d),abs(abs(x%d)-d))<<endl;
      }else{
        cout<<max(abs(x%d),abs(abs(x%d)-d))<<endl;
      }
    }
  }
}