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 <vector>
#include <cmath>

using namespace std;

#define EPS 1e-10

struct Point {
	double x, y;

	Point() {}
	Point(double x, double y) :x(x), y(y) {}

	Point operator+(const Point& p) const { return Point(x + p.x, y + p.y); }
	Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
	Point operator*(const double& k) const { return Point(x * k, y * k); }
	Point operator/(const double& k) const { return Point(x / k, y / k); }

	friend istream& operator>>(istream& is, Point& p) {
		is >> p.x >> p.y;
		return is;
	}
};

typedef Point Vector;

double norm(Vector a) { return a.x * a.x + a.y * a.y; }
double abs(Vector a) { return sqrt(norm(a)); }

struct Circle {
	Point c;
	double r;

	Circle() {}
	Circle(Point c, double r) :c(c), r(r) {}
};

double arg(Vector p) { return atan2(p.y, p.x); }
Point polar(double a, double r) { return Point(cos(r) * a, sin(r) * a); }

vector<Point> getCrossPointCC(Circle c1, Circle c2) {
	double d = abs(c1.c - c2.c);
	double a = acos((c1.r * c1.r + d * d - c2.r * c2.r) / (2 * c1.r * d));
	double t = arg(c2.c - c1.c);
	vector<Point> ps;
	ps.push_back(c1.c + polar(c1.r, t + a)); ps.push_back(c1.c + polar(c1.r, t - a));
	return ps;
}

int main() {
	int N, K; cin >> N >> K;
	vector<Point> ps(N);
	vector<double> c(N);
	for (int i = 0; i < N; ++i) cin >> ps[i] >> c[i];
	double l = 0, r = 1e9;
	for (int _ = 0; _ < 300; ++_) {
		double mid = (l + r) / 2;
		bool flag = false;
		for (int i = 0; i < N; ++i) {
			int cnt = 0;
			for (int j = 0; j < N; ++j)
				if (abs(ps[i] - ps[j]) <= mid / c[j] + EPS) ++cnt;
			if (cnt >= K) flag = true;
		}
		for (int i = 0; i < N; ++i) {
			Circle A(ps[i], mid / c[i]);
			for (int j = i + 1; j < N; ++j) {
				Circle B(ps[j], mid / c[j]);
				if (abs(A.c - B.c) >= A.r + B.r) continue;
				if (abs(A.c - B.c) < abs(A.r - B.r)) continue;
				auto ccp = getCrossPointCC(A, B);
				int cnt1 = 0, cnt2 = 0;
				for (int k = 0; k < N; ++k) {
					if (abs(ccp[0] - ps[k]) <= mid / c[k] + EPS) ++cnt1;
					if (abs(ccp[1] - ps[k]) <= mid / c[k] + EPS) ++cnt2;
				}
				if (cnt1 >= K || cnt2 >= K) flag = true;
			}
		}
		if (flag) r = mid;
		else l = mid;
	}
	printf("%.10f\n", r);

	return 0;
}