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)

// Hail god Yato
 
#include <bits/stdc++.h> 
using namespace std;
 
#define hs ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long ll;
const ll mod = 1000000007;
const ll INF = 1e18;
const ll MAX = 100001;
//
//
void solve(){
    ll n, k, d;
    cin>>n>>k>>d;
    // ll sign = -1*(ans<0);
    n = abs(n);
    if(abs(n) > k*d){
        ll ans = abs(n) - k*d;
        cout<<ans;
        return ;
    }
    ll times = n/d;
    ll pos = n-times*d;
    if((k-times)&1)
        cout<<abs(pos-d);
    else
        cout<<pos;

}
int main(){ 
        hs;
        ll t;
        t=1;
        // cin>>t;
        for (int i=1; i<=t; i++){
                //cout<<"Case #"<<i<<": ";
                solve();
         }
        return 0; 
}