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>
using namespace std;
using namespace __detail;

typedef long long ll;
#define ar array
#define vi vector<int>
#define vvi vector<vector<int>>
#define vll vector<long long>
#define all(x) (x).begin(), (x).end()
#define en '\n'
#define FILL(x, v) memset(x, v, sizeof(x))
#define pb push_back
#define fast                     \
	ios::sync_with_stdio(false); \
	cin.tie(NULL);               \
	cout.tie(NULL);
#define F_OR(i, a, b, s) for (int i = (a); ((s) > 0 ? i < (b) : i > (b)); i += (s))
#define F_OR1(e) F_OR(i, 0, e, 1)
#define F_OR2(i, e) F_OR(i, 0, e, 1)
#define F_OR3(i, b, e) F_OR(i, b, e, 1)
#define F_OR4(i, b, e, s) F_OR(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)
#define FOR(...)       \
	F_ORC(__VA_ARGS__) \
	(__VA_ARGS__)

#define E_ACH2(x, a) for (auto &x : a)
#define E_ACH3(x, y, a) for (auto &[x, y] : a)
#define E_ACH4(x, y, z, a) for (auto &[x, y, z] : a)
#define E_ACHC(...) GET5(__VA_ARGS__, E_ACH4, E_ACH3, E_ACH2)
#define EACH(...)       \
	E_ACHC(__VA_ARGS__) \
	(__VA_ARGS__)

template <typename A, typename B>
inline void chmin(A &a, B b)
{
	if (a > b)
		a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b)
{
	if (a < b)
		a = b;
}

int main()
{
	fast;
	ll x, k, d;
	cin >> x >> k >> d;
	if (x)
	{
		while (k && x > 0)
		{
			x -= d;
			k--;
		}
	}
	else
	{
		while (k && x < 0)
		{
			x += d;
			k--;
		}
	}
	if (x < 0)
	{
		if (k & 1)
			cout << abs(x + d) << en;
		else
			cout << abs(x) << en;
	}
	else if (x > 0)
	{
		if (k & 1)
			cout << abs(x - d) << en;
		else
			cout << abs(x) << en;
	}
	else
	{
		if (k & 1)
			cout << abs(d) << en;
		else
			cout << abs(x) << en;
		return 0;
	}
}