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<iostream>
#include<math.h>
#include<vector>
#include<array>
#include<algorithm>
#include<numeric>
#include<map>
#include<queue>
#include<deque>
#include<set>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
typedef vector<double> vec;

int N,K;
vec x(100),y(100),c(100);
bool b(double t){
    int rev=0;
    for(int i=0;i<N;i++){
        int include=0;
        for(int j=0;j<N;j++){
            if(j==i){
                include++;
                continue;
            }
            double d=sqrt((x[j]-x[i])*(x[j]-x[i])+(y[j]-y[i])*(y[j]-y[i]));
            if(t/c[j]-t/c[i]>d) include++;
        }
        rev=max(rev,include);
    }
    for(int i=0;i<N;i++){
        for(int j=i+1;j<N;j++){
            double d=sqrt((x[j]-x[i])*(x[j]-x[i])+(y[j]-y[i])*(y[j]-y[i]));
            double ri=t/c[i],rj=t/c[j];
            if(ri+rj<d) continue;
            vec v={(x[j]-x[i])/d,(y[j]-y[i])/d};
            vec w={(y[j]-y[i])/d,-(x[j]-x[i])/d};
            double X=(d+(ri*ri-rj*rj)/d)/2.0;
            double h=sqrt(ri*ri-X*X);
            int count=0;
            vec L={x[i]+X*v[0]+h*w[0],y[i]+X*v[1]+h*w[1]};
            for(int k=0;k<N;k++){
                if(k==i || k==j)count++;
                else if(sqrt((x[k]-L[0])*(x[k]-L[0])+(y[k]-L[1])*(y[k]-L[1]))<t/c[k])count++;
            }
            rev=max(rev,count);
            count=0;
            L={x[i]+X*v[0]-h*w[0],y[i]+X*v[1]-h*w[1]};
            for(int k=0;k<N;k++){
                if(k==i || k==j)count++;
                else if(sqrt((x[k]-L[0])*(x[k]-L[0])+(y[k]-L[1])*(y[k]-L[1]))<t/c[k])count++;
            }   
            rev=max(rev,count);         
        }
    }
    return rev>=K;
}
double BS(double l,double r){
    if(r-l<1.0e-10)return r;
    double m=(l+r)/2.0;
    if(b(m)) return BS(l,m);
    else     return BS(m,r);
}

int main(){
    scanf("%d %d",&N,&K);
    for(int i=0;i<N;i++){
        scanf("%lf %lf %lf",&x[i],&y[i],&c[i]);
    }
    printf("%.10lf\n",BS(0.0,1.0e10));
    return 0;
}