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<bits/stdc++.h>
#include <ctime>
using namespace std;
#define ll long long
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define N 101
#define mod 1000000007
#define f(i,n) for(int i=0;i<n;i++)
#define in(i,arr) for(auto &i:arr)cin>>i;
#define all(x) (x).begin(), (x).end()   // sort(all(v)) instead sort(v.begin(),v.end())
int binary(vector<int> arr,int ele,int beg=1)
{   int l=arr.size()-1,mid;
    while(beg<=l)
    {
        mid=(beg+l)/2;
        if(arr[mid]==ele)
            return mid;
        else if(arr[mid]>ele)
            l=mid-1;
        else
            beg=mid+1;
    }
    return -1;

}
ll powe(ll base,ll pow)
{
    ll res=1;
    while(pow>0)
    {
        if(pow&1)
            res*=base;
        else
            base*=base,pow/=2;
    }
    return res;
}
int main() {
    fast;
    int ti = 1;
    //cin >> ti;

    while (ti--){
        ll n,k,d;
        cin>>n>>k>>d;
        if(d==abs(n))
        {
            if(k&1)
                cout<<0;
            else
                cout<<abs(n);
            continue;
        }
        else if(d<abs(n)) {
            ll ans = abs(n) % d;
            ll moves= abs(n)/d;
            if(moves==k)
                cout<<ans;
            else if(moves>k)
            {
                cout<<abs(n)-k*d;
            }
            else {
                k-=moves;
                if(k&1)
                    cout<<d-ans;
                else cout<<ans;
            }

        }
        else {
            ll ans=abs(n)-d;
            if(k&1)
                cout<<abs(ans);
            else
                cout<<abs(n);
        }

    }

}