#include <bits/stdc++.h>
using namespace std;
int N, K; 
double X[64], Y[64], C[64]; 
vector<pair<double, double>> P;
double eps = 1e-10; 
void add1(double x1, double x2, double y1, double y2, double r1, double r2){
	double a = x2-x1; double b = y2-y1; 
	double k = (r1*r1-r2*r2+a*a+b*b)*0.5; 
	double dis = (a*a+b*b)*r1*r1-k*k; 
	if(dis < -eps) return; 
	else if(abs(dis) <= eps){
		double x = k*a / (a*a + b*b); double y = k*b / (a*a + b*b); 
		P.push_back(make_pair(x+x1, y+y1)); 
	}
	else{
		double xp = (k*a + b * pow(dis, 0.5)) / (a*a + b*b); 
		double yp = (k*b - a * pow(dis, 0.5)) / (a*a + b*b); 
		double xm = (k*a - b * pow(dis, 0.5)) / (a*a + b*b); 
		double ym = (k*b + a * pow(dis, 0.5)) / (a*a + b*b); 
		P.push_back(make_pair(xp+x1, yp+y1)); P.push_back(make_pair(xm+x1, ym+y1)); 
	}
}
void add2(double a, double b){
	P.push_back(make_pair(a, b)); 
}
bool check(double t, vector<pair<double, double>> &P){
	int mx = 0; 
	for(auto p : P){
		int cnt = 0; 
		for(int i = 0; i < N; i++){
			double dx = p.first - X[i]; 
			double dy = p.second - Y[i]; 
			if(pow(dx*dx + dy*dy, 0.5) <= t/C[i] + eps) cnt ++; 
		}
		mx = max(mx, cnt); 
	}
	return mx >= K; 
}
int main(){
	cin >> N >> K; 
	for(int i = 0; i < N; i++){
		cin >> X[i] >> Y[i] >> C[i]; 
	}
	if(K == 1){
		cout << 0 << endl; return 0; 
	}
	double ok = 314514.1919; double ng = 0; 
	for(int t = 0; t < 100; t++){
		double mid = (ok + ng) * 0.5; 
		for(int i = 0; i < N; i++){
			for(int j = i+1; j < N; j++){
				add1(X[i], X[j], Y[i], Y[j], mid/C[i], mid/C[j]); 
			}
		}
		for(int i = 0; i < N; i++) add2(X[i], Y[i]); 
		if(check(mid, P)) ok = mid; 
		else ng = mid; 
		P.clear(); 
	}
	cout.precision(15);
	cout << ok << endl;  
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
using namespace std;
int N, K; 
double X[64], Y[64], C[64]; 
vector<pair<double, double>> P;
double eps = 1e-10; 
void add1(double x1, double x2, double y1, double y2, double r1, double r2){
	double a = x2-x1; double b = y2-y1; 
	double k = (r1*r1-r2*r2+a*a+b*b)*0.5; 
	double dis = (a*a+b*b)*r1*r1-k*k; 
	if(dis < -eps) return; 
	else if(abs(dis) <= eps){
		double x = k*a / (a*a + b*b); double y = k*b / (a*a + b*b); 
		P.push_back(make_pair(x+x1, y+y1)); 
	}
	else{
		double xp = (k*a + b * pow(dis, 0.5)) / (a*a + b*b); 
		double yp = (k*b - a * pow(dis, 0.5)) / (a*a + b*b); 
		double xm = (k*a - b * pow(dis, 0.5)) / (a*a + b*b); 
		double ym = (k*b + a * pow(dis, 0.5)) / (a*a + b*b); 
		P.push_back(make_pair(xp+x1, yp+y1)); P.push_back(make_pair(xm+x1, ym+y1)); 
	}
}
void add2(double a, double b){
	P.push_back(make_pair(a, b)); 
}
bool check(double t, vector<pair<double, double>> &P){
	int mx = 0; 
	for(auto p : P){
		int cnt = 0; 
		for(int i = 0; i < N; i++){
			double dx = p.first - X[i]; 
			double dy = p.second - Y[i]; 
			if(pow(dx*dx + dy*dy, 0.5) <= t/C[i] + eps) cnt ++; 
		}
		mx = max(mx, cnt); 
	}
	return mx >= K; 
}
int main(){
	cin >> N >> K; 
	for(int i = 0; i < N; i++){
		cin >> X[i] >> Y[i] >> C[i]; 
	}
	if(K == 1){
		cout << 0 << endl; return 0; 
	}
	double ok = 314514.1919; double ng = 0; 
	for(int t = 0; t < 100; t++){
		double mid = (ok + ng) * 0.5; 
		for(int i = 0; i < N; i++){
			for(int j = i+1; j < N; j++){
				add1(X[i], X[j], Y[i], Y[j], mid/C[i], mid/C[j]); 
			}
		}
		for(int i = 0; i < N; i++) add2(X[i], Y[i]); 
		if(check(mid, P)) ok = mid; 
		else ng = mid; 
		P.clear(); 
	}
	cout.precision(15);
	cout << ok << endl;  
}
```