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>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstring>
#include <chrono>
#include <complex>
#define REP(i,a,b) for (auto i = a; i != b; i++)
#define ll long long int
#define ld long double
#define vi vector<int>
#define vll vector<ll>
#define vvi vector < vi >
#define pii pair<int,int>
#define pll pair<long long, long long>
#define mod 1000000007
#define inf 1000000000000000001
#define all(c) c.begin(),c.end()
#define mp(x,y) make_pair(x,y)
#define mem(a,val) memset(a,val,sizeof(a))
#define eb emplace_back
#define f first
#define s second
#define pb push_back
#define SQ(a) (a)*(a)
 
using namespace std;


int main()
{
    std::ios::sync_with_stdio(false);
    int T = 1;
    //cin>>T;
    for(int t = 1;t<=T;t++)
    {
       	ll x,k,d;
       	cin>>x>>k>>d;
       	ll val1 = abs(x/d);
       	//cout<<val1<<endl;
       	if(k > val1)
       	{
       		ll y = val1*d;
       		k = k - val1;
       		if(k % 2 == 0)
       			cout<<min(abs(x - y),abs(x+y))<<endl;
       		else 
       			cout<<min(min(abs(x - y - d),abs(x - y + d)),min(abs(x + y + d),abs(x + y - d)))<<endl;
       	}
       	else
       	{
       		cout<<min(abs(x - k*d),abs(x + k*d))<<endl;
       	}
       	
    }

    return 0;
}