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>
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
typedef long long ll;
typedef long double ld;
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define mod 1000000007
#define inf 1000000000000000000
#define bpc(x) __builtin_popcountll(x)
#define autoit(x,it) for(auto it = x.begin(); it != x.end(); it++)
#define rep(n) for(ll i = 0; i < n; i++)
#define repi(i,n) for(ll i = 0; i < n; i++)

#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 

#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> 

using namespace std;

int main()
{
  FAST/**/
   
   ll x,k,d;
   cin>>x>>k>>d;
   
   ll ans = LLONG_MAX;
   ans = min({ans,abs(x+k*d),abs(x-k*d)});
   ll req = abs(x)/d;
   for(ll i=req-100;i<=req+100;i++)
   {
       if(i>=0 && i%2 == k%2 && i<=k)
       {
            ans = min({ans,abs(x+i*d),abs(x-i*d)});
       }
   }
   
   cout<<ans;
   
   return 0;        
}