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)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
#define ll          long long
#define pb          push_back
#define INF         9223372036854775807
#define    endl        '\n'
#define pii         pair<ll int,ll int>
#define vi          vector<ll int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll int)x.size()
#define hell        1000000007
#define rep(i,a,b)    for(ll int i=a;i<b;i++)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
#define lower(u)    transform(u.begin(), u.end(), u.begin(), ::tolower);//convert string u to lowercase;
#define upper(u)    transform(u.begin(), u.end(), u.begin(), ::toupper);
using namespace std;

using namespace __gnu_pbds; 
  
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
#define N  100005

void solve()
{
    ll x,k,d;
    cin>>x>>k>>d;
    x=abs(x);
    ll t=x/d;
    if(k<=t)
    {
        ll ans=x-k*d;
        cout<<ans<<endl;
    }
    else
    {
        k=k-t;
        ll ans=x-t*d;
        if(k%2==0)
        {
            cout<<ans<<endl;
        }
        else
        {
            ans=abs(ans-d);
            cout<<ans<<endl;
        }
    }
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    int TESTS=1;
//    cin>>TESTS;
    while(TESTS--)
    {
        solve();
    }
    return 0;
}