Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

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

Sample Output 1

2.4

If we put the heat source at \left(-0.2, 0\right), the 1-st, 2-nd, and 3-rd pieces of meat will be ready to eat within 2.4 seconds. This is the optimal place to put the heat source.

Sample Input 2

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

Sample Output 2

7411.2252

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// F.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;

int n, k;
vector<double> x, y, c;

struct Point {
	double x;
	double y;
	double dist(const Point& p) {
		double dx = x - p.x, dy = y - p.y;
		return sqrt(dx * dx + dy * dy);
	}
};

struct Circle {
	Point p;
	double c;
	double r;
	vector<Point> cross(const Circle& c) {
		vector<Point> pp;
		double dist = p.dist(c.p);
		if (dist <= r + c.r && r <= dist + c.r && r + dist >= c.r) {
			double dx = c.p.x - p.x, dy = c.p.y - p.y;
			double a = (dx * dx + dy * dy + r * r - c.r * c.r) / 2, dd = dx * dx + dy * dy;
			pp.push_back({
				(a * dx + dy * sqrt(dd * r * r - a * a)) / dd + p.x,
				(a * dy - dx * sqrt(dd * r * r - a * a)) / dd + p.y
			});
			pp.push_back({
				(a * dx - dy * sqrt(dd * r * r - a * a)) / dd + p.x,
				(a * dy + dx * sqrt(dd * r * r - a * a)) / dd + p.y
			});
		}
		return pp;
	}
};

int main(int argc, char* argv[]) {
	cout.precision(20);
#ifdef _MSC_VER
	while (true)
#endif
	{
		n = -1;
		cin >> n >> k;
		if (n < 0) return 0;
		x.resize(n);
		y.resize(n);
		c.resize(n);
		vector<Circle> cc(n);
		for (int i = 0; i < n; ++i) {
			cin >> x[i] >> y[i] >> c[i];
			cc[i].p.x = x[i];
			cc[i].p.y = y[i];
			cc[i].c = c[i];
		}
		double left = 0, right = 1e9;
		for (int t = 0; t < 100; ++t) {
			double mid = (left + right) / 2;
			for (int i = 0; i < n; ++i) {
				cc[i].r = mid / cc[i].c;
			}
			vector<Point> pp;
			for (int i = 0; i < n; ++i) {
				pp.push_back(cc[i].p);
				for (int j = i + 1; j < n; ++j) {
					for (auto p : cc[i].cross(cc[j])) {
						pp.push_back(p);
					}
				}
			}
			int mc = 0;
			for (auto p : pp) {
				int cnt = 0;
				for (auto c : cc) {
					cnt += p.dist(c.p) <= c.r + 1e-12;
				}
				mc = max(mc, cnt);
			}
			if (mc >= k) {
				right = mid;
			} else {
				left = mid;
			}
		}
		cout << right << endl;
	}
	return 0;
}