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;


int main()
{
    long long int x,k,d;
    cin>>x>>k>>d;
    if(x==0)
    {
        if(k%2==0)
        {
            cout<<0<<endl;
        }
        else
        {
            cout<<x+d<<endl;
        }
    }
    else
    {
        if(x>0)
        {
            long long int n= x/d+1;
            if(k>n)
            {
                long long int a1 = x +(n-1)*1LL*(d)*(-1);
                long long int a2 = x+(n)*1LL*d*(-1);
                long long int m = min(abs(a1),abs(a2));
                cout<<m<<endl;
            }
            else
            {
                cout<<abs(x-k*d)<<endl;
            }
        }
        else
        {
            long long int n = (x*(-1))/d+1;
          
            if(k>n)
            {
                long long int a1 = x+(n-1)*1LL*(d);
                long long int a2 = x+(n)*1LL*d;
                long long int m= min(abs(a1),abs(a2));
                cout<<m<<endl;
            }
            else
            cout<<abs(x+k*1LL*d)<<endl;
        }
    }
}