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 <iostream>
#include <iomanip>
#include<math.h>
#include<list>
#include <algorithm>
#include<set>
#include<vector>
#include<math.h>
#include<map>
#include<string>
#include <numeric>
#include <queue>
#include <sstream>
#include <bitset>
#include<stack>

using namespace std;
using ll = long long;
using vll = vector<long long>;
using sll = set<long long>;
const long long ll_max = 9223372036854775807;
const int int_max = 2147483647;

template<typename T>
T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<typename T>
T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T>
map<T, T> getPrimeFactor(T n) {
	map<T, T> res;
	for (T i = 2; i * i <= n; ++i) {
		while (n % i == 0) {
			res[i] ++;
			n /= i;
		}
	}
	if (n != 1) res[n] = 1;
	return res;
}

template<typename T>
bool IsPrimeNumber(T num)
{
	if (num <= 2) return true;
	else if (num % 2 == 0) return false;

	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			return false;
		}
	}

	return true;
}

long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}

// 繰り返し二乗法
ll pow2(ll n, ll p, ll mod) {
	if (p == 0 || n == 1) {
		return 1;
	}
	else {
		ll ret = pow2(n * n % mod, p / 2, mod);
		if (p % 2 == 1) {
			ret *= n;
		}
		return ret % mod;
	}
}


#define rep(i,s,e) for(ll i=s;i<e;i++)
#define repeq(i,s,e) for(ll i=s;i<=e;i++)

int main() {
	ll X, K, D;
	cin >> X >> K >> D;

	if (X < 0) {
		X = -X;
	}

	ll k = X / D;
	ll ans = X % D;

	if (X == 0) {
		k = 0;
		ans = X;
	}

	K -= k;
	if (K < 0) {
		ans -= D * K;
	}
	else if (K & 1) {
		ans -= D;
	}

	cout << abs(ans) << endl;
	return 0;

}