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 rep(i,n) for(int i=0;i<(n);i++)
#define PI acos(-1)
typedef  long long ll;
ll gcd(ll x,ll y){
   if(y==0) return x;
   else return gcd(y,x%y);
}
ll lcm(ll x,ll y){
   return x/gcd(x,y)*y;
}

int main(){
   ll x,k,d;
   cin>>x>>k>>d;
   ll ans=10000000000000000;
      ll a=-x/(2*d)+k/2;

   for(ll i=a-10000;i<a+10000;i++){
      if(i<0){
         continue;
      }
      ll y=x+(2*i-k)*d;
      ll z=abs(y);
      if((y-x)/d!=(2*i-k)){
         continue;
      }
      ans=min(ans,z);
   }  
   
   cout<<ans<<endl;
   return 0;
}