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 <iostream>
#include <algorithm>
#include <functional>
#include <cmath>
using namespace std;

int main(void){
    long long int X,K,D;
    cin>>X>>K>>D;
  		
  	int logX = log10(X);
  	int logK = log10(K);
  	int logD = log10(D);
  	
  	long long int answer = 0;
  
  	if(logD+logK<15){
	  	if(abs(X)>abs(D*K)){
  	        answer = abs(X) - abs(D*K);
        	cout<< answer<<endl;
          	return 0;
        }
    }
  	
  	long long int n = X/D;
  	long long int min = X%D;
  	
  	
  	if( (K-n)%2 == 0){
      	answer = abs(min);
    }else{
      	answer = abs(D)-abs(min);
    }
  	
  	cout<<answer<<endl;
  	return 0;
}