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;
#define ALL(obj) (obj).begin(),(obj).end()
#define SORTD(s) sort((s).rbegin(),(s).rend())
#define rep(i,n) for(int i=0;i<(n);i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
typedef long long ll;
#define PI 3.14159265358979323846264338327950L
const int mod=1000000007,MAX=200005,INF=1<<30;

int main() {
	ll x,k,d;
  	cin>>x>>k>>d;
  	
  	if(x<0)x*=-1;
  	ll ans=0;
  	ll i=0;
  	ll xp=x;
  
  	if(x/d>k){
    	ans=xp-(d*k);
    }
  
  	else{
      	ll y;
		i=x/d;
      	x-=y*d;
      	k-=y;
      
      	ll j=i-1;
      	ll op=abs(x-(d*i));
        ll pp=abs(x-(d*j));
      
      	if(k%2==0){
        	ans=op;
        }
      
      	else{
        	ans=pp;
        }
    }
  
  cout<<ans<<endl;
}