#include <bits/stdc++.h>
#define Int int64_t

using namespace std;

vector<pair<double, double>> crosspoint(double x1, double y1, double r1, double x2, double y2, double r2) {
	x1 -= x2;
	y1 -= y2;
	double S = x1 * x1 + y1 * y1;
	double a = (S + r2 * r2 - r1 * r1) / 2.0;
	double D = S * r2 * r2 - 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 main() {
	const double EPS = 1e-9;
	int N, K;
	cin >> N >> K;
	vector<double> x(N), y(N), c(N);
	for (int i = 0; i < N; ++i) {
		cin >> x[i] >> y[i] >> c[i];
	}

	auto check = [&](double mid) {
		vector<double> r(N);
		for (int i = 0; i < N; ++i) { r[i] = mid / c[i]; }
		vector<pair<double, double>> ps;
		for (int i = 0; i < N; ++i) {
			ps.emplace_back(x[i], y[i]);
			for (int j = i + 1; j < N; ++j) {
				auto vec = crosspoint(x[i], y[i], r[i], x[j], y[j], r[j]);
				for (auto& p : vec) { ps.push_back(p); }
			}
		}
		for (auto& p : ps) {
			int cnt = 0;
			double X = p.first, Y = p.second;
			for (int i = 0; i < N; ++i) {
				if (r[i] + EPS >= hypot(X - x[i], Y - y[i])) { ++cnt; }
			}
			if (cnt >= K) { return true; }
		}
		return false;
	};

	double ok = 1e9, ng = 0;
	int loop = 0;
	while (loop++ < 100) {
		double mid = (ok + ng) / 2;
		(check(mid) ? ok : ng) = mid;
	}
	cout.precision(15);
	cout << fixed << ok << endl;

	return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
#define Int int64_t

using namespace std;

vector<pair<double, double>> crosspoint(double x1, double y1, double r1, double x2, double y2, double r2) {
	x1 -= x2;
	y1 -= y2;
	double S = x1 * x1 + y1 * y1;
	double a = (S + r2 * r2 - r1 * r1) / 2.0;
	double D = S * r2 * r2 - 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 main() {
	const double EPS = 1e-9;
	int N, K;
	cin >> N >> K;
	vector<double> x(N), y(N), c(N);
	for (int i = 0; i < N; ++i) {
		cin >> x[i] >> y[i] >> c[i];
	}

	auto check = [&](double mid) {
		vector<double> r(N);
		for (int i = 0; i < N; ++i) { r[i] = mid / c[i]; }
		vector<pair<double, double>> ps;
		for (int i = 0; i < N; ++i) {
			ps.emplace_back(x[i], y[i]);
			for (int j = i + 1; j < N; ++j) {
				auto vec = crosspoint(x[i], y[i], r[i], x[j], y[j], r[j]);
				for (auto& p : vec) { ps.push_back(p); }
			}
		}
		for (auto& p : ps) {
			int cnt = 0;
			double X = p.first, Y = p.second;
			for (int i = 0; i < N; ++i) {
				if (r[i] + EPS >= hypot(X - x[i], Y - y[i])) { ++cnt; }
			}
			if (cnt >= K) { return true; }
		}
		return false;
	};

	double ok = 1e9, ng = 0;
	int loop = 0;
	while (loop++ < 100) {
		double mid = (ok + ng) / 2;
		(check(mid) ? ok : ng) = mid;
	}
	cout.precision(15);
	cout << fixed << ok << endl;

	return 0;
}

```