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>   
#define ll long long int  
#define mkp make_pair 
#define mod 1000000007 
#define maxn 100001 
#define ff first
#define ss second
using namespace std;     
bool check(int i,int j,int k){  
    return (i!=j&&j!=k&&k!=i);
    } 

int main(){    
  ios_base::sync_with_stdio(false); 
  cin.tie(NULL);    
  ll x,k,d; 
  cin>>x>>k>>d;  
  ll temp=x/d; 
  if(k<=temp){   
     if(x>=0)
     cout<<abs(x-(k*d))<<"\n"; 
     else 
     cout<<abs(x+(k*d))<<"\n";
  } 
  else{  
     k-=temp;  
    if(x>=0){
     if(k&1) 
    cout<<min(abs(x-(temp*d)-d),abs(x-(temp*d)+d))<<"\n"; 
      else  
       cout<<(ll)abs(x-(temp*d))<<"\n"; 
    } 
   else {  if(k&1)
         cout<<min(abs(x+(temp*d)-d),abs(x+(temp*d)+d))<<"\n"; 
      else  
       cout<<(ll)abs(x+(temp*d))<<"\n"; 
     }
    }
 
       
return 0;  
}