Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 3
-1 0 3
0 0 3
1 0 2
1 1 40

Sample Output 1

2.4

If we put the heat source at \left(-0.2, 0\right), the 1-st, 2-nd, and 3-rd pieces of meat will be ready to eat within 2.4 seconds. This is the optimal place to put the heat source.

Sample Input 2

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

Sample Output 2

7411.2252

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

/* by Natsu Kinmoe */
#include <bits/stdc++.h>
using namespace std;

#define all(x) (x).begin(),(x).end()
#define loop(i,n) for(int i=0;i<(n);i++)
#define cont(i,n) for(int i=1;i<=(n);i++)
#define circ(i,a,b) for(int i=(a);i<=(b);i++)
#define range(i,a,b,c) for(int i=(a);((c)>0?i<=(b):i>=(b));i+=(c))
#define foreach(it,v) for(__typeof((v).begin()) it=(v).begin(),_e_D_=(v).end();it!=_e_D_;it++)
#define y0 y0O0OO00OO0OO0OO0OOO00OO0OO0O0O000OO0
#define y1 y1II11II11III11I1III11II111IIII1II1I1
#define pub push_back
#define pob pop_back
#define mak make_pair

typedef long long ll;
typedef long double lf;
const int Inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3fll;
/* Source code starts here */

int n,k;
int x[66],y[66],c[66];
lf ds[66];

lf dist(lf x,lf y){
	return sqrtl(x*x+y*y);
}

int main(){
	scanf("%d%d",&n,&k);
	cont(i,n)scanf("%d%d%d",x+i,y+i,c+i);
	if(k==1)return 0&puts("0.0000000000");
	lf l=0,r=1000000;
	loop(tms,150){
		lf m=(l+r)/2;bool can=0;
		cont(i,n)ds[i]=m/c[i];
		cont(j,n)cont(i,j){
			lf dis=dist(x[i]-x[j],y[i]-y[j]);
			lf xi,yi;
			if(dis>=ds[i]+ds[j])continue;
			if(ds[i]>ds[j]+dis)xi=dis,yi=0;
			else if(ds[j]>ds[i]+dis)xi=0,yi=0;
			else{
				xi=(ds[i]*ds[i]-ds[j]*ds[j]+dis*dis)/(2*dis);
				yi=sqrt(ds[i]*ds[i]-xi*xi);
			}
			lf xx=(x[j]-x[i])/dis,yy=(y[j]-y[i])/dis;
			range(tm,-1,1,2){
				lf nx=xi*xx+yi*yy*tm+x[i];
				lf ny=xi*yy-yi*xx*tm+y[i];
				int cnt=0;
				cont(i,n)if(dist(x[i]-nx,y[i]-ny)<=ds[i]+1e-10)cnt++;
				if(cnt>=k)can=1;
			}
		}
		if(can)r=m;
		else l=m;
	}
	printf("%.10Lf\n",l);
	return 0;
}