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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int 
#define ld long double
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define pi 3.14159265358979323846
#define mod 1000000007
int main()
{ 
    ll p;
    p=1;
    while(p--)
    {
        ll a,b,c,d,e,cnt=0,x,k,y;
        cin>>x>>k>>d;
        x=abs(x);
        a=x/d;
        if(a==k)
        {
            x=x-a*d;
            cout<<abs(x)<<endl;
        }
        else if(a>k)
        {
            x=x-k*d;
            cout<<abs(x)<<endl;
        }
        else
        {
            x=x-a*d;
            y=x-d;
            if((k-a)%2==0)
            cout<<abs(x)<<endl;
            else
            cout<<abs(y)<<endl;
        }
    }
}