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 long long ll;
typedef long double ld;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;

typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<vi> vvi;

 #define rep(i, a, b) for (int i=a; i<(b); i++)
    
 #define sz(x) (int)(x).size()
 #define mp make_pair
 #define pb push_back
 #define f first
 #define s second
 #define lb lower_bound
 #define ub upper_bound
 #define all(x) x.begin(), x.end()
 

 #define trav(container, it) for(typeof(container.begin()) it=container.begin(); it!=container.end(); it++)
 
 #define PI 3.14159265358979323846264338327950L
const int MOD = 1000000007;
const char nl = '\n';
 
int main() 
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    ll x,k,d;cin>>x>>k>>d;

    ll res = x%d;

    if(k>=(abs(x-res)/d))
    {
        k-=(abs(x-res)/d);

        if(k%2 == 0)
        {
            cout<<res;
        }
        else cout<<abs(res-d);
    }
    else
    {
        cout<<min(abs(x-k*d),abs(x+k*d));
    }

}