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>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
using P = pair<ll, ll>;
using V = vector<ll>;
#define rep(i, n) for(ll i=0;i<(ll)(n);i++)
#define rep1(i, n) for(ll i=1;i<=(ll)(n);i++)
#define rep2(i, m, n) for(ll i=m;i<(ll)(n);i++)
#define rrep(i, n, m) for(ll i=n;i>=(ll)(m);i--)
#define ALL(obj) (obj).begin(), (obj).end()
#define RALL(obj) (obj).rbegin(), (obj).rend()
#define MOD 1000000007
#define INF 1000000000
#define LLINF 1e18

void solve_abc_a() {
	ll n, m;
	ll x, y;
	ll a, b, d;
	ll c;
	//char c;
	ll h, w;
	ll k;
	string s, t;
	ll l_ans = 0;
	string s_ans = "NG";

	cin >> x >> k >> d;

	ll cnt = abs(x) / d;
	if (cnt >= k) {
		l_ans = abs(x) - k * d;
	}
	else {
		if ((k - cnt) % 2 == 0) {
			l_ans = abs(x) - d * cnt;
		}
		else {
			l_ans = abs(x) - d * (cnt + 1);
		}
	}

	cout << abs(l_ans) << "\n";

	return;
}


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed;
	cout << setprecision(7);

	solve_abc_a();

	return 0;
}