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 <bits/stdc++.h>
using namespace std;

int n, k;
double x[60], y[60], c[60], r[60];
double EPS = 1e-9;

double dist(pair<double, double> p, pair<double, double> q) {
	double xx = (p.first - q.first)*(p.first - q.first);
	double yy = (p.second - q.second)*(p.second - q.second);

	return sqrt(xx + yy);
}

vector<pair<double, double>> circle_circle_intersection(double x1, double y1, double r2, double x2, double y2, double r1){
	x1 -= x2;
	y1 -= y2;
	double S = x1*x1 + y1*y1;
	double a = (S + r1*r1 - r2*r2) / 2;
	double D = S*r1*r1-a*a;
	if(D <= 0) return {};
 
	double A1 = a * x1;
	double B1 = y1 * sqrt(D);
	double A2 = a * y1;
	double B2 = x1 * sqrt(D);
	return {
		{(A1+B1)/S+x2, (A2-B2)/S+y2},
		{(A1-B1)/S+x2, (A2+B2)/S+y2}
	};
}

int check(double t) {
	int ans = 0;
	for (int i = 0; i < n; i++) {
		r[i] = t/c[i];
	}

	for (int i = 0; i < n; i++) {
		pair<double, double> p(x[i], y[i]);
		int cnt = 0;
		for (int k = 0; k < n; k++) {
			if (dist(p, make_pair(x[k], y[k])) <= r[k]) cnt++;
		}
		ans = max(ans, cnt);
	}

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			vector<pair<double, double>> pts = circle_circle_intersection(x[i], y[i], r[i], x[j], y[j], r[j]);
			for (auto p: pts) {
				int cnt = 0;
				//cout << i << ' ' << j << ' ' << p.first << ' ' << p.second << '\n';
				//cout << x[i] << ' ' << y[i] << ' ' << r[i] << '\n';
				//cout << x[j] << ' ' << y[j] << ' ' << r[j] << '\n';
				for (int k = 0; k < n; k++) {
					if (dist(p, make_pair(x[k], y[k])) <= r[k] + EPS) cnt++;
				}
				ans = max(ans, cnt);
			}
		}
	}

	return ans;
}

signed main() {
	ios::sync_with_stdio(0); cin.tie(0);
	feenableexcept(29);

	cin >> n >> k;
	if (k == 1) {
		cout << "0\n"; exit(0);
	}
	for (int i = 0; i < n; i++) {
		cin >> x[i] >> y[i] >> c[i];
	}

	double lo = 0, hi = 100000000;
	for (int ite = 0; ite < 100; ite++) {
		double mid = (lo + hi)/2;

		if (check(mid) >= k) {
			hi = mid;
		} else {
			lo = mid;
		}
	}
	//cout << check(hi) << '\n';
	cout << fixed << setprecision(9) << hi << '\n'; 

	return 0;
}