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)

//ssenseeeeeeeeeeeeeeeeeeeeeeeeeeeee
#include<bits/stdc++.h>
using namespace std;
typedef long long  ll;
typedef unsigned long long ull;
#define MX 1000000000
#define endl '\n'
#define startt ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fr first
#define sc second
#define MOD 1000000007

int main()
{
    startt
    ll x, k, d;
    cin >> x >> k >> d;
    x = abs(x);
    if(k*d < x)
    {
        cout << x-k*d;
        return 0;
    }
    ull bigger = x%d;
    ull smaller = d-bigger;
    ull bigp = k-x/d;
    if(bigp%2 == 0)
    {
        cout << bigger << endl;
    }
    else
    {
        cout << smaller << endl;
    }
}