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>
using namespace std;

typedef pair<int,int> P;
#define ll long long
#define ld long double
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define PI 3.14159265358979323846
#define sz(x) ((int)(x).size())
#define chmin(x,y) x = min(x,y)
#define chmax(x,y) x = max(x,y)
#define all(x) (x).begin(),(x).end()

const int INF = 1e9;
const int MOD = 1e9 + 7;
const ll LINF = 1e18;
int main()
{
    ll x,k,d;
    cin >> x >> k >> d;
    x = abs(x);
    if((x-k*d)>=0)
    {
        cout << x-k*d << endl;
        return 0;
    }else
    {
        if((k-(x/d))%2==0)
        {
            cout << abs(x-d*(x/d)) << endl;
            return 0;
        }else
        {
            cout << abs(x-d*((x/d)+1)) << endl;
            return 0;
        }
        
    }
    

}