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>
/*...
  ...
  author 
  @
  Aman Kothari
  .....
  .....*/
using namespace std;
#define ll long long
#define dl double

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
	ll x,d,k;
    	cin>>x>>k>>d;
    	vector<ll> v;
    	if(x==k && k==d){
    		cout<<x<<"\n";
		}
		else{
			if(x>=0){
			ll	prev=x-d;
    	while(k--){
    		if(abs(x-d)<=abs(x+d)){
    		x-=d;
					}
			else{
				x+=d;
			}	
			prev=abs(x);
//			cout<<prev<<"\n";	
    		v.push_back(abs(x));
		}
		cout<<*min_element(v.begin(), v.end())<<"\n"; 
					}
			else{
				ll prev=x+d;
				while(k--){
    	if(abs(x+d)<=abs(x-d)){
    		x+=d;
					}
			else{
				x-=d;
			}	
			prev=abs(x);
//			cout<<prev<<"\n";
    		v.push_back(abs(x));
		}
		cout<<*min_element(v.begin(), v.end())<<"\n"; 
			}		
				}
    return 0;
}