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)


/*
	*	Created By: 'Present_Sir'
	*	Created On: Sunday 16 August 2020 12:24:48 PM IST
*/

#include<bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL)
#define int long long
using namespace std;


void solve(){
    int x,k,d;
    cin>>x>>k>>d;

    int t = (x+d-1)/d;
    //cout<<t<<endl;
    if(t <= k ){
        int pos = min({abs(x-t*d),abs(x+t*d)});
        if( (k-t)%2 == 0) cout<<pos<<endl;
        else{
            int mn = abs(pos+d);
            mn = min(mn,abs(pos-d));
            cout<<mn<<endl;
        }
    }else{
        cout<<min({abs(x-k*d),abs(x+k*d)})<<endl;
    }
}

int32_t main(){
    IOS;
    //int t; cin>>t; while(t--)
    {
        solve();
    }
    return 0;
}