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<cstdio>
#include<cmath>
#include<limits>
#include<string.h>
#include<vector>
#include<stack>
#include<queue>
using namespace std;

int main()
{
    int64_t x,k,d;
    cin>>x>>k>>d;
    int64_t a,b;
    if (x>=0)
    {
        if (x/d>=k)
        {
            cout<<x-(k*d)<<endl;
        }else
        {
            if (k%2==0)
            {
                a=x%(2*d);
                b=(2*d)-a;
                cout<<min(b,a)<<endl;               
            }else
            {
                a=(x+d)%(2*d);
                b=(2*d)-a;
                cout<<min(b,a)<<endl;                  
            }
            
            


        }
        
        
    }else
    {
        if (k<=(-x)/d)
        {
            cout<<(-x-(k*d))<<endl;
        }else
        {
            if (k%2==0)
            {
                a=(-x)%(2*d);
                b=(2*d)-a;
                cout<<min(b,a)<<endl;                
            }else
            {
                a=(-x+d)%(2*d);
                b=(2*d)-a;
                cout<<min(b,a)<<endl;   
            }
            
            
            

        }
        
        
    }
    
    

    return 0;
}