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>
//Fast io
#define io ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
//Class shorthands
#define ll long long
#define db long double
#define pii pair<int,int>
#define vi vector<int>
#define vii vector<pii>
#define mi map<int,int>
#define mii map<pii,int>
//Function Shorthands
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define pb push_back
#define mp make_pair
#define po pop_back
#define eb emplace_back
//Code Shorthands
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repe(i,a,b) for(int i=a;i<=b;i++)
#define repr(i,a,b) for(int i=b;i>=a;i--)
#define all(a) (a).begin(),(a).end()
#define sz(x) (int)x.size()
//Constants
#define hell 1000000007
#define PI acos(-1.0)
using namespace std;
bool check(ll a, ll b, ll c){
	int m = max({a,b,c});
	if(a==b || a==c || b==c) return false;
	return a+b+c-m>m;
}
void solve(){
	ll x,k,d;
	cin>>x>>k>>d;
	x = max(x, -1*x);
	ll f = x/d;
	if(f>k){
		cout<<x-k*d<<endl;
	}
	else{
		k-=f;
		if(k%2==0) cout<<x-f*d<<endl;
		else cout<<-1*(x-(f+1)*d)<<endl;
	}
}

int main(){
	io;
	int t = 1;
	//cin>>t;
	while(t--){
		solve();
	}
}