#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <vector>
using namespace std;

int N, K, X[66], Y[66], C[66]; double R[66];

bool chk(double r)
{
	for (int i = 0; i < N; i++) R[i] = r / C[i];

	double per = 2 * acos(-1.0);
	for (int i = 0; i < N; i++){
		int base = 1;
		vector<pair<double, int> > stop;
		for (int j = 0; j < N; j++) if (i != j){
			double dist = hypot(X[i] - X[j], Y[i] - Y[j]);
			if (dist >= R[i] + R[j]) continue;
			if (R[i] >= R[j] && dist <= R[i] - R[j]) continue;
			if (R[i] <= R[j] && dist <= R[j] - R[i]){ base++; continue; }

			double dir = atan2(Y[j] - Y[i], X[j] - X[i]);
			double rng = acos((R[i] * R[i] + dist * dist - R[j] * R[j]) / (2 * R[i] * dist));
			double st = dir - rng, ed = dir + rng;

			auto norm = [per](double &x){
				x = fmod(x, per);
				if (x < 0) x += per;
			};
			norm(st);
			norm(ed);

			stop.push_back({ st, +1 });
			if (ed < st){
				stop.push_back({ per, -1 });
				stop.push_back({ 0, +1 });
			}
			stop.push_back({ ed, -1 });
		}

		if (base >= K) return true;
		sort(stop.begin(), stop.end());
		for (auto &p : stop){
			base += p.second;
			if (base >= K) return true;
		}
	}
	return false;
}

int main()
{
	scanf ("%d %d", &N, &K);
	for (int i = 0; i < N; i++) scanf ("%d %d %d", &X[i], &Y[i], &C[i]);

	double l = 0, r = 1e6;
	for (int k = 0; k < 100; k++){
		double m = (l + r) / 2;
		if (chk(m)) r = m;
		else l = m;
	}

	printf ("%.12lf\n", r);
	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 <stdio.h>
#include <math.h>
#include <algorithm>
#include <vector>
using namespace std;

int N, K, X[66], Y[66], C[66]; double R[66];

bool chk(double r)
{
	for (int i = 0; i < N; i++) R[i] = r / C[i];

	double per = 2 * acos(-1.0);
	for (int i = 0; i < N; i++){
		int base = 1;
		vector<pair<double, int> > stop;
		for (int j = 0; j < N; j++) if (i != j){
			double dist = hypot(X[i] - X[j], Y[i] - Y[j]);
			if (dist >= R[i] + R[j]) continue;
			if (R[i] >= R[j] && dist <= R[i] - R[j]) continue;
			if (R[i] <= R[j] && dist <= R[j] - R[i]){ base++; continue; }

			double dir = atan2(Y[j] - Y[i], X[j] - X[i]);
			double rng = acos((R[i] * R[i] + dist * dist - R[j] * R[j]) / (2 * R[i] * dist));
			double st = dir - rng, ed = dir + rng;

			auto norm = [per](double &x){
				x = fmod(x, per);
				if (x < 0) x += per;
			};
			norm(st);
			norm(ed);

			stop.push_back({ st, +1 });
			if (ed < st){
				stop.push_back({ per, -1 });
				stop.push_back({ 0, +1 });
			}
			stop.push_back({ ed, -1 });
		}

		if (base >= K) return true;
		sort(stop.begin(), stop.end());
		for (auto &p : stop){
			base += p.second;
			if (base >= K) return true;
		}
	}
	return false;
}

int main()
{
	scanf ("%d %d", &N, &K);
	for (int i = 0; i < N; i++) scanf ("%d %d %d", &X[i], &Y[i], &C[i]);

	double l = 0, r = 1e6;
	for (int k = 0; k < 100; k++){
		double m = (l + r) / 2;
		if (chk(m)) r = m;
		else l = m;
	}

	printf ("%.12lf\n", r);
	return 0;
}
```