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)

// first second push_back unordered return continue break vector visited check flag bool while iterator begin end lower_bound upper_bound temp true false ll_MAX ll_MIN insert erase clear pop push compare ll64_MAX ll64_MIN  reverse replace stringstream string::npos length substr front priority_queue

#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#include <bits/stdc++.h>
using namespace std;
#define debug(...) 42
#endif

#define endl '\n'
#define ll long long
#define rint register int
#define minpq priority_queue <int, vector<int>, greater<int> >
#define maxpq  priority_queue <int>

#define re register
#define pb(x) push_back(x);
#define ce(x) cout << x << '\n';

using db = long double;
using pll = pair < ll, ll >;
#define scan(a, n) 		for(int i = 0; i < n; i++)cin >> a[i];

#define rep(i,x,n,inc) for(i=x ; i<n ; i+=inc)
#define repr(i,x,n,inc) for( i=x ; i>n ; i+=inc)
#define all(a)      (a).begin(),(a).end()
#define unique_sort(x) sort(all(x)), x.resize(distance(x.begin(), unique(all(x))))

#define mp(a,b) make_pair(a,b)
#define ff first
#define ss second
#define hell 1000000007
#define infl LLONG_MAX

#define conutBits(n)	__builtin_popcount(n)
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
#define Fill(s,v) memset(s,v,sizeof(s))
#define cout_p(x, p) cout << fixed << setprecision((p)) << x << endl     //print with precision

#define tc(tt) \
    ll tt;    \
    cin >> tt; \
    for (ll _tt = 0; _tt < tt; _tt++) // testcase 

int32_t main() {
	ios::sync_with_stdio(0); 		cin.tie(0); cout.tie(0);

	ll t, i, x, j, y, d, 	z, k, n;
	cin >> x >> k >> d;

	x = abs(x);
	z = x;

	t = x / d;
	t = min(t, k);
	k -= t;
	x -= d * t;

	if (k) {
		if (k & 1) {
			z = min({z, abs(x - d), abs(x + d)});
			if (t & 1)
				z = min({z, abs(x)});
		} else {
			z = min(z, abs(x));
		}
	}
	else
		z = min(z, abs(x));

	ce(z);
	debug(z, x);
}