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++ (Clang 10.0.0) (judge verdict: Accepted)

/*
    Name:zhangchuanyan
    Copyright:HHTC
    Author:zhangchuanyan
    Date: 2020/8/15 20:53:15
    Description:
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
#define ll long long
using namespace std;
int main()
{
	ll x,k,d;
	cin>>x>>k>>d;
	x=abs(x);
	if(x==0){
		if(k%2==0){
			cout<<"0\n";
		}else 
		    cout<<d<<endl;
	}else {
	    ll num=x/d;
		if(num>=k){
			cout<<x-k*d<<endl;
		}else{
			
			ll kk=k-num;
			if(kk%2==0){
				cout<<abs(x-num*d)<<endl;
			}else{
				cout<<abs(x-num*d-d)<<endl;
			}
		}
	}
    return 0;
}