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>
#include<algorithm>

using namespace std;
typedef long long int l;
#define mod 1000000007
#define t1 l t; cin>>t; while(t--)
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL)
#define vl vector<l>
#define pb push_back
#define ub upper_bound
#define lb lower_bound

int main() {
fast;
l x,k,d;
cin>>x>>k>>d;
if(x==d)
{
    if(k%2==0)
    cout<<x;
    else
    cout<<0;
}else{
    l a=x/d;
    if(a>=k)
    cout<<x-(k*d);
    else{
        l b=x-a*d;
        l c=b-d;
       /* if(k-a==1)
        cout<<abs(c);*/

            if(k%2==0)
            {
                if(a%2==0)
                cout<<abs(b);
                else
                cout<<abs(c);
            }else{
                if(a%2==0)
                cout<<abs(c);
                else
                cout<<abs(b);
            }
        }
    }



	return 0;
}