#include<bits/stdc++.h>
int main(){
    int n,K;std::cin>>n>>K;
    std::vector<std::complex<double>>P(n);
    std::vector<double>C(n),R(n);
    for(int i=0;i<n;i++){
        double x,y;std::cin>>x>>y;
        P[i]={x,y};
        std::cin>>C[i];
    }
    double eps=1e-12;
    double lwr=0,upr=200000;
    for(int ngtkana=0;ngtkana<70;ngtkana++){
        double T=(lwr+upr)/2;
        for(int i=0;i<n;i++)R[i]=T/C[i];
        auto covered_by_K=[&](std::complex<double>p, int i, int j){
            int counter=0;
            for(int k=0;k<n;k++){
                counter+=k==i||k==j||std::abs(p-P[k])<R[k];
            }
            return K<=counter;
        };
        for(int i=0;i<n;i++)if(covered_by_K(P[i],i,i))goto Upr;
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                double d=std::abs(P[j]-P[i]);
                if(d<std::abs(R[j]-R[i])+eps||R[i]+R[j]-eps<d)continue;
                double t0=std::arg(P[j]-P[i]);
                double t1=std::acos((R[i]*R[i]+d*d-R[j]*R[j])/(2*R[i]*d));
                if(covered_by_K(P[i]+R[i]*std::exp(std::complex<double>(0,t0+t1)),i,j)
                    ||covered_by_K(P[i]+R[i]*std::exp(std::complex<double>(0,t0-t1)),i,j)){
                    goto Upr;
                }
            }
        }
        lwr=T;
        continue;
        Upr:upr=T;
    }
    std::cout<<std::setprecision(13)<<lwr<<std::endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include<bits/stdc++.h>
int main(){
    int n,K;std::cin>>n>>K;
    std::vector<std::complex<double>>P(n);
    std::vector<double>C(n),R(n);
    for(int i=0;i<n;i++){
        double x,y;std::cin>>x>>y;
        P[i]={x,y};
        std::cin>>C[i];
    }
    double eps=1e-12;
    double lwr=0,upr=200000;
    for(int ngtkana=0;ngtkana<70;ngtkana++){
        double T=(lwr+upr)/2;
        for(int i=0;i<n;i++)R[i]=T/C[i];
        auto covered_by_K=[&](std::complex<double>p, int i, int j){
            int counter=0;
            for(int k=0;k<n;k++){
                counter+=k==i||k==j||std::abs(p-P[k])<R[k];
            }
            return K<=counter;
        };
        for(int i=0;i<n;i++)if(covered_by_K(P[i],i,i))goto Upr;
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                double d=std::abs(P[j]-P[i]);
                if(d<std::abs(R[j]-R[i])+eps||R[i]+R[j]-eps<d)continue;
                double t0=std::arg(P[j]-P[i]);
                double t1=std::acos((R[i]*R[i]+d*d-R[j]*R[j])/(2*R[i]*d));
                if(covered_by_K(P[i]+R[i]*std::exp(std::complex<double>(0,t0+t1)),i,j)
                    ||covered_by_K(P[i]+R[i]*std::exp(std::complex<double>(0,t0-t1)),i,j)){
                    goto Upr;
                }
            }
        }
        lwr=T;
        continue;
        Upr:upr=T;
    }
    std::cout<<std::setprecision(13)<<lwr<<std::endl;
}

```