#include<bits/stdc++.h>
#define eps 5e-7
using namespace std;
int n,k;
double x[62],y[62],c[62];
queue<tuple<double,double,double> > q;
double xp,yp,ap,lb;
double opt(double xx,double yy,double aa){
	vector<double> aux;
	for(int i=1;i<=n;i++){
		aux.push_back(c[i]*hypot(max(abs(xx-x[i])-aa,0.00),max(abs(yy-y[i])-aa,0.00)));
	}
	nth_element(aux.begin(),aux.begin()+k-1,aux.end());
	return aux[k-1];
}
double rel(double xx,double yy){
	vector<double> aux;
	for(int i=1;i<=n;i++){
		aux.push_back(c[i]*hypot(xx-x[i],yy-y[i]));
	}
	nth_element(aux.begin(),aux.begin()+k-1,aux.end());
	return aux[k-1];
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>k;
	for(int i=1;i<=n;i++)
		cin>>x[i]>>y[i]>>c[i];
	q.emplace(0.0,0.0,1000.0);
	double res=1e9;
	while(q.size()){
		tie(xp,yp,ap)=q.front();
		q.pop();
		lb=opt(xp,yp,ap);
		if((res-lb)<=(eps*max(1.0,lb)))
			continue;
		res=min(res,rel(xp,yp));
		for(double dx:{-0.5,0.5}){
			for(double dy:{-0.5,0.5}){
				q.emplace(xp+dx*ap,yp+dy*ap,ap/2.00);
			}
		}
	}
	cout<<fixed<<setprecision(15)<<res<<endl;
	return 0;
}
### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
#define eps 5e-7
using namespace std;
int n,k;
double x[62],y[62],c[62];
queue<tuple<double,double,double> > q;
double xp,yp,ap,lb;
double opt(double xx,double yy,double aa){
	vector<double> aux;
	for(int i=1;i<=n;i++){
		aux.push_back(c[i]*hypot(max(abs(xx-x[i])-aa,0.00),max(abs(yy-y[i])-aa,0.00)));
	}
	nth_element(aux.begin(),aux.begin()+k-1,aux.end());
	return aux[k-1];
}
double rel(double xx,double yy){
	vector<double> aux;
	for(int i=1;i<=n;i++){
		aux.push_back(c[i]*hypot(xx-x[i],yy-y[i]));
	}
	nth_element(aux.begin(),aux.begin()+k-1,aux.end());
	return aux[k-1];
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>k;
	for(int i=1;i<=n;i++)
		cin>>x[i]>>y[i]>>c[i];
	q.emplace(0.0,0.0,1000.0);
	double res=1e9;
	while(q.size()){
		tie(xp,yp,ap)=q.front();
		q.pop();
		lb=opt(xp,yp,ap);
		if((res-lb)<=(eps*max(1.0,lb)))
			continue;
		res=min(res,rel(xp,yp));
		for(double dx:{-0.5,0.5}){
			for(double dy:{-0.5,0.5}){
				q.emplace(xp+dx*ap,yp+dy*ap,ap/2.00);
			}
		}
	}
	cout<<fixed<<setprecision(15)<<res<<endl;
	return 0;
}
```