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>
using namespace std;
typedef long double ld;
const int N = 1e2+5;
const ld eps = 1e-10;

struct P{
	ld x,y;
	P(){x = 0;y = 0;}
	P(ld _x,ld _y){x = _x;y = _y;}
	ld dis(){return sqrt(x*x+y*y);}
	P norm(){ld t = (*this).dis();return P(x/t,y/t);}
	P rot(){return P(-y,x);}
//	P add(P t){return P(x+t.x,y+t.y);}
//	P sub(P t){return P(x-t.x,y-t.y);}
//	P mul(ld t){return P(x*t,y*t);}
	P operator + (const P& t){return P(this -> x+t.x,this -> y+t.y);}
	P operator - (const P& t){return P(this -> x-t.x,this -> y-t.y);}
	P operator * (const ld& t){return P(this -> x*t,this -> y*t);}
	
} p[N];
ld c[N],r[N];
int n,k;

inline bool OK(P now){
	int res = 0;
	for(int i = 0;i < n;++i)
		if((now-p[i]).dis() < r[i]+eps) res++;
	return res >= k;
}

inline bool get_P(P &p1,P &p2,int i,int j){
	ld dv = (p[i]-p[j]).dis();
    if (dv-abs(r[i]-r[j]) < eps) return 0;
    if (r[i]+r[j]-dv < eps) return 0;
 
    ld nv = r[i]*r[i]-r[j]*r[j];
    ld pv = nv/dv;
    ld av = (dv+pv)/2;
 
    P dir = (p[j]-p[i]).norm();
    P pg = (p[i]+dir*av);
 
    ld ov = sqrt(r[i]*r[i]-av*av);
    p1 = (pg+dir.rot()*ov);
    p2 = (pg+dir.rot()*(-ov));
	return 1;
}

inline bool check(ld T){
	for(int i = 0;i < n;++i) r[i] = T/c[i];
	for(int i = 0;i < n;++i)
		if(OK(p[i])) return 1;
	
	P p1,p2;
	for(int i = 0;i < n;++i){
		for(int j = i+1;j < n;++j){
			if(get_P(p1,p2,i,j)){
				if(OK(p1) || OK(p2)) return 1;
			}
		}
	}
	return 0;
}

inline ld bin_search(ld l,ld r){
	if(k == 1) return 0.0;
	for(int i = 0;i < 100;++i){
		ld m = (l+r)/2.0;
		if(check(m)) r = m;
		else l = m;
	}
	return l;
}

int main()
{
	std::ios::sync_with_stdio(0);cin.tie(0);
	
	cin >> n >> k;
	for(int i = 0;i < n;++i){
		cin >> p[i].x >> p[i].y >> c[i];
	}
	printf("%.8Lf\n",bin_search(0,1e6));
	return 0;
}