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 <complex>
#include <vector>
using namespace std;

//  |p-o1|=r1, |p-o2|=r2 を満たす2個の点を返す
vector<complex<double>> circle_intersection(complex<double> o1, double r1, complex<double> o2, double r2)
{
    double d = abs(o1-o2);
    double cos = (r1*r1+d*d-r2*r2)/(2*r1*d);
    if (cos<-1.0 || 1.0<cos)
        return vector<complex<double>>();
    complex<double> p(r1*cos, r1*sqrt(1-cos*cos));
    vector<complex<double>> r{p, conj(p)};
    for (complex<double> &t: r)
        t = t*(o2-o1)/abs(o2-o1)+o1;
    return r;
}

#include <iostream>
#include <cmath>

int main()
{
    int N, K;
    cin>>N>>K;
    vector<complex<double>> p(N);
    vector<double> c(N);
    for (int i=0; i<N; i++)
    {
        double x, y;
        cin>>x>>y>>c[i];
        p[i] = complex<double>(x, y);
    }

    auto check = [&](complex<double> o, double t)
    {
        int k = 0;
        for (int i=0; i<N; i++)
            if (c[i]*abs(o-p[i])<=t)
                k++;
        return k>=K;
    };

    double l = 0.;
    double r = 1000000.;
    for (int i=0; i<100; i++)
    {
        double m = (l+r)/2.;
        bool ok = false;
        for (auto p1: p)
            if (check(p1, m))
                ok = true;
        for (int j=0; j<N; j++)
            for (int k=0; k<N; k++)
                for (complex<double> x: circle_intersection(p[j], m/c[j]-1e-8, p[k], m/c[k]-1e-8))
                    if (check(x, m))
                        ok = true;
        (ok ? r : l) = m;
    }
    printf("%.10f\n", l);
}