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

Solution (in CPP):

Solution:
/*
读题不规范，爆零两行泪。
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
#define urd uniform_real_distribution
mt19937 rng(19260817);
double sq(double x){return x*x;}
const int N=60;
int n,m;
double x[N+1],y[N+1],c[N+1];
double a[N+1];
double calc(double x0,double y0){
	for(int i=1;i<=n;i++)a[i]=c[i]*sqrt(sq(x[i]-x0)+sq(y[i]-y0));
	sort(a+1,a+n+1);
	return a[m];
}
double sim_ann(double st,double ed,double delta){
	double x0=0,y0=0,res=calc(0,0);
//	int cnt=0;
	for(double tem=st;tem>=ed;tem*=delta){
		double nw_x=x0+(urd<>(0,10000)(rng)-urd<>(0,10000)(rng))*tem;
		double nw_y=y0+(urd<>(0,10000)(rng)-urd<>(0,10000)(rng))*tem;
		double nw=calc(nw_x,nw_y);
		if(nw<res||urd<>(0,1)(rng)<=exp((res-nw)/tem))x0=nw_x,y0=nw_y,res=nw;
//		if((++cnt)%100==0)printf("tem=%.5lf x=%.5lf y=%.5lf res=%.5lf\n",tem,x0,y0,res);
	}
	return res;
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=n;i++)cin>>x[i]>>y[i]>>c[i];
	double ans=sim_ann(1e4,1e-10,.9999);
	rng=mt19937(114514);
	ans=min(ans,sim_ann(1e4,1e-10,.9999));
	printf("%.10lf",ans);
	return 0;
}