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>
using namespace std;
#define ll long long int
#define vll vector<ll>
#define fast  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

ll solve(ll x,ll k,ll d)
{
    ll ans,i,j,n;
     if(x!=0)
     {
         x=abs(x);
         if((x-(k*d))>=0)
            ans=x-(k*d);
         else
         {
             n=x/d;
             x=x-(n*d);
             k=k-n;
             if(k%2==0)
                ans=x;
             else ans=min(abs(x-d),abs(x+d));
         }
     }
     else{
        if(k%2==0)
            ans=0;
        else ans=x+d;
     }
     return ans;
}
int main()
{
    ll t=1;
    ll x,k,d,ans;
    while(t--)
    {
        cin>>x>>k>>d;
        ans=solve(x,k,d);
        cout<<ans<<endl;
    }
}