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>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_poizcy.hpp>
typedef long long  ll;
#define pb                push_back
#define mp                make_pair
#define all(a)            (a).begin(), (a).end()
#define clr(a,h)          memset(a, (h), sizeof(a))
#define mem(a,h)          memset(a, (h), sizeof(a))
#define fi first
#define se second
#define por(a,b) (((a%MOD) * (b%MOD))%MOD)
#define forg(i, b, e, c)    for (ll i = (ll)b; i < (ll)e; i+=c)
#define forr(i, b, e)    for (ll i = b; i < e; i++)
 
 
using namespace std;
//using namespace __gnu_pbds;
typedef double lldb;
typedef pair<ll, ll>  ii;
typedef pair<double, double>  iidb;
typedef pair<ll, ii>  iii;
typedef vector<ll>     vi;
typedef vector<vi>     vvi;
typedef vector<ii>      vii;
typedef vector<ll>      vll;
//typedef tree<ii,null_type,less<ii>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
const ll INF = 1e9;
const double PI = acos(-1);
const ll MOD= 200003;
#define initseg ll new_nodo=(nodo*2),mid=(iz+der)/2;
 
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define tam 2100000
#define offset 200
#define ptr nodo*

int main()
{
	ll valor,k,d;
	cin>>valor>>k>>d;
	if ((valor/d)+1<=k)
	{
		

		if (ll(valor/d)%2!=k%2)
			valor=(valor-ll(valor/d)*d-d);
		else
			valor-=ll(valor/d)*d;
		valor=abs(valor);
		
	}
	else
	{
		valor-=k*d;
	}
	cout<<valor<<endl;
}