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:
#include<bits/stdc++.h>
//#include<iostream>
//#include<cstdio>
//#include<cstring>
//#include<string>
//#include<queue>
//#include<cmath>
//#include<algorithm>
//#define debug1 cout<<"?"<<endl;
//#define debug2 cout<<"?"<<i<<endl;
#define PII pair<int,int>
typedef long long ll;
const int maxn=3e5+9;
const double pi=acos(-1);
const int mod=998244353;
const int INF=0x3f3f3f3f;
const double ep=1e-7;
using namespace std;
struct point{
	double x,y,c,r;
}p[maxn],p1[maxn];
int n,k,tot;
int sgn(double x)
{
	if(fabs(x)<ep)return 0;
	if(x>0)return 1;return -1;
}
inline double len2(point a,point b)
{
	return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
inline void getCircleIntersection(point a,point b)
{
	double ab,ae,h;
	ab=len2(a,b);//圆心距
	if(sgn(ab)==0)return;//同心圆
	ae=(a.r*a.r-b.r*b.r+ab)/2;
	h=ab*a.r*a.r-ae*ae;
	if(h<0)return;
	h=sqrt(h);
	p1[++tot].x=a.x+ae*(b.x-a.x)/ab-(b.y-a.y)*h/ab;
	p1[tot].y=a.y+ae*(b.y-a.y)/ab+(b.x-a.x)*h/ab;
	p1[++tot].x=a.x+ae*(b.x-a.x)/ab+(b.y-a.y)*h/ab;
	p1[tot].y=a.y+ae*(b.y-a.y)/ab-(b.x-a.x)*h/ab;
}
inline bool solve(double r)
{
	for(int i=1;i<=n;i++)p[i].r=r/p[i].c;
	for(int i=1;i<=n;i++)
	{
		p1[++tot].x=p[i].x,p1[tot].y=p[i].y;
		for(int j=1;j<i;j++)getCircleIntersection(p[i],p[j]);
	}
	for(int i=1;i<=tot;i++)
	{
		int sum=0;
		for(int j=1;j<=n;j++)
		{
			if(sgn(sqrt(len2(p1[i],p[j]))*p[j].c-r)<=0)sum++;
		}
		if(sum>=k)return 1;
	}
	return 0;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
	scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].c);
	double l=0,r=100000000;
	while(sgn(l-r)<0)
	{
		tot=0;
		double mid=(l+r)/2;
		if(solve(mid))r=mid;
		else l=mid;
	}
	printf("%f\n",r);
}