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[65], y[65], c[65], r[65];
double eps = 1e-8;

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 r1, double x2, double y2, double r2) {
	vector<pair<double, double> > ans;
	double d = dist(make_pair(x1, y1), make_pair(x2, y2));
	if (d > r1 + r2) {
		return ans;
	}
	if (x1 == x2 && y1 == y2) {
		ans.push_back(make_pair(x1, y1)); return ans;
	}
	if (d < abs(r1 - r2)) {
		ans.push_back(make_pair(x1, y1)); return ans;
	}

	double a = (r1*r1 - r2*r2 + d*d)/(2*d);
	double h = sqrt(r1*r1 - a*a);


	double x3 = (x2 - x1)*(a/d) + x1;
	double y3 = (y2 - y1)*(a/d) + y1;

	ans.emplace_back(x3 + h*(y2 - y1)/d, y3 - h*(x2 - x1)/d);
	ans.emplace_back(x3 - h*(y2 - y1)/d, y3 + h*(x2 - x1)/d);

	return ans;
}

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++) {
		for (int j = i; 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;
				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;
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	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 l = 0, r = 100000000;
	while(r-l>=1e-8)
	{
		double mid = (l + r)/2;

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

	return 0;
}