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 <iostream>
#include <string.h>
#include <algorithm>
#include <cstring>
#include <math.h>
#define ll long long
ll gcd(ll a,ll b){
    ll s=1,c;
    while(a&&b){
        if((~a&1)&&(~b&1))
            a>>=1,b>>=1,s<<=1;
        else if(~a&1)a>>=1;
        else if(~b&1)b>>=1;
        else if(a>b)a=a-b;
        else c=b-a,b=a,a=c;
    }
    if(!a)return b*s;
    if(!b)return a*s;
}
using namespace std;
int main(){
	ll int x,k,d,ans;
	cin>>x>>k>>d;
	x=abs(x);
	if(k*d<x)
		ans=x-k*d;
	else {
			ans=abs(x%d-abs(k-x/d)%2*d);
		}
	cout<<ans;
	return 0;
}