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>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define all(cont) cont.begin(), cont.end()
#define foreach(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define fore(i,a,b) for(int i=a,almo5t=b;i<almo5t;++i)
#define SZ(x) ((int)x.size())
#define EPS 1e-9
#define PI acos(-1)
#define MOD 1000000007
#define FIN std::ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

const int MAXN = 1e7;

typedef long long int ll;

using namespace std;

ll x,k,d;

ll ts()
{
    if(x>=k*d){
        return x-k*d;
    }else{
        k -= x/d;
        //cout<<(x%d)-d<<" "<<x%d<<" "<<k<<" \n";
        if((k>=2)){
            return min(x%d,abs((x%d)-d));
        }else{
            return abs((x%d)-d);
        }
    }
}

int main(){
    FIN;
    //freopen("../../input.txt", "r", stdin);freopen("../../output.txt", "w", stdout);
    cin>>x>>k>>d;
    x = abs(x);
    ll ans = ts();
    cout<<ans<<"\n";
    return 0;
}

// ============== Notes =============== //
// common bugs
// - remove reader 
// - fastio
// - int overflow (especially intermediate expressionns)
// - array bounds (indices bigger than MAXN?, is MAXN correct?)
// - special cases (n=1, graph not connected?)
// - re-initialize efficiently between test cases