Problem: Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.

Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.

Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.

Constraints

* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100

Input

Input is given from Standard Input in the following format:


N K
x_1 y_1 c_1
\vdots
x_N y_N c_N


Output

Print the answer.

It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.

Examples

Input

4 3
-1 0 3
0 0 3
1 0 2
1 1 40


Output

2.4


Input

10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16


Output

7411.2252

Solution (in CPP):

Solution:
#include <iostream>
#include <cmath>
#include <complex>
#include <string>
#include <sstream>
#include <limits>
#include <algorithm>
#include <functional>

#include <vector>
#include <queue>

using namespace std;

int N, K;
vector<tuple<int, int, int>> info;

double nth_distance(double x, double y)
{
	vector<double> dist(N);
	for (int i = 0; i < N; i++)
	{
		dist[i] = get<2>(info[i]) * hypot(x - get<0>(info[i]), y - get<1>(info[i]));
	}

	nth_element(dist.begin(), dist.begin() + K - 1, dist.end());
	return dist[K - 1];
}

double nth_distance_lb(double x, double y, double r)
{
	vector<double> dist(N);
	for (int i = 0; i < N; i++)
	{
		dist[i] = get<2>(info[i]) * hypot(max(abs(x - get<0>(info[i])) - r, 0.0), max(abs(y - get<1>(info[i])) - r, 0.0));
	}

	nth_element(dist.begin(), dist.begin() + K - 1, dist.end());
	return dist[K - 1];
}

const double eps = 1e-6;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

	cin >> N >> K;

	for (int i = 0; i < N; i++)
	{
		int x, y, c;
		cin >> x >> y >> c;

		info.emplace_back(x, y, c);
	}

	double result = 1e9;

	queue<tuple<double, double, double>> q;
	q.emplace(0, 0, 1000);
	while (!q.empty())
	{
		double x, y, r;
		tie(x, y, r) = q.front();
		q.pop();

		double lb = nth_distance_lb(x, y, r);
		if (result - lb > eps * max(1.0, lb))
		{
			result = min(result, nth_distance(x, y));

			for (double dx : {-0.5, 0.5})
			{
				for (double dy : {-0.5, 0.5})
				{
					q.emplace(x + dx * r, y + dy * r, r / 2.0);
				}
			}
		}
	}

	cout.setf(ios::fixed);
	cout.precision(15);
	cout << result;

	return 0;
}