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 unsigned long long ull;
#define max3(a,b,c) max(a,max(b,c))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))
#define min3(a,b,c) min(a,min(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define yes cout<<"YES"<<nl
#define no  cout<<"NO"<<nl
#define nl  '\n'
#define tr ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define tst int t;cin>>t;while(t--)
#define     Max             1000000000000000014
#define     Min             -1000000000000000014
#define     PI              3.1415926535897
#define     Fr               first
#define     Se               second
#define     all(v)          v.begin(),v.end()
#define     mem(a,v)        memset(a,v,sizeof(a))
#define     space            ' '
#define COUT(v) for(auto it:v)cout<<it<<space;cout<<nl;
/*error cheaking method starts from here */
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }

void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
cerr << *it << " = " << a << endl;
err(++it, args...);
}
/*error cheaking method ebds here */

#define vi    vector<int>
#define    vl    vector<ll>
#define    si    set<int>
#define    sl    set<ll>
#define    msi    multiset<int>
#define    msl    multiset<ll>
#define    pii    pair<int,int>
#define    pil    pair<int,ll>
#define    pli    pair<ll,int>
#define    pll    pair<ll,ll>
#define    mp    make_pair
#define    pb    push_back
int main()
{
    ll x,k,d;
    cin>>x>>k>>d;
   ll m=x/d;
   if(m==1&&x==d)
   {
       if(k%2==0)
            cout<<x;
       else
        cout<<0;
   }
   else
   {
       if(m<k)
       {
           x-=(m*d);k-=m;

       }
       else
       {
           x-=(k*d);k=0;
       }
       cout<<x;
   }
}