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)

/*
___ is a raging sadist.
*/
//Destroy all WA at test 2
// Stay cool
// Pass the 4th wall..

#include <bits/stdc++.h>
#define fastio()                  \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0)
#define pb push_back
#define show(x) cout << (#x) << " : " << x << endl;
#define ll long long
#define ld long double
#define pow power
#define mp make_pair
#define ff first
#define ss second
#define pii pair<ll, ll>
#define sq(x) ((x) * (x))
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define siz(a) int((a).size())
#define For(i,a,b) for(int (i)=(a);(i) < (b); ++(i))
#define endl "\n"
#define pi 3.14159265
const ll mod = 1000 * 1000 * 1000 + 7;
const ll mod1 = 998244353;
const ll INF     = 1ll*1000*1000*1000*1000*1000*1000 + 7; 
using namespace std;

ll power(ll x, ll y) 
{ 
   ll res = 1;      
  
   
    while (y > 0) 
    { 
       
        if (y & 1) 
            res = (long long)(res*x); 
  

        y = y>>1; 

        if (x<=100000000)
        x = (long long)(x*x);

        //cout<<x<<'\n'; 
    } 
    return res; 
} 






int main()
{

#ifndef ONLINE_JUDGE    
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif

fastio();

    ll x,k,d;
    cin>>x>>k>>d;

    ll q=abs(x)/d;

    if (q>=k)
    {
        cout<<abs(x)-k*d;

    } else {

        if (x<=0)
        {
            ll x1=x+q*d;
            ll x2=x+(q+1)*d;
            k-=q;

            if (k&1)
                cout<<abs(x2);
            else
                cout<<abs(x1);

        } else {
            ll x1=x-q*d;
            ll x2=x-(q+1)*d;
            k-=q;

            if (k&1)
                cout<<abs(x2);
            else
                cout<<abs(x1);
        }
    }

return 0;
}