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;

vector<pair<double, double>> crosspoint(double x1, double y1, double r2, double x2, double y2, double r1){
    x1 -= x2;
    y1 -= y2;
    double S = x1*x1 + y1*y1;
    double a = (S + r1*r1 - r2*r2) / 2;
    double D = S*r1*r1-a*a;
    if(D < 0) return {};

    double A1 = a * x1;
    double B1 = y1 * sqrt(D);
    double A2 = a * y1;
    double B2 = x1 * sqrt(D);
    return {
        {(A1+B1)/S+x2, (A2-B2)/S+y2},
        {(A1-B1)/S+x2, (A2+B2)/S+y2}
    };
}

double EPS = 1e-9;

int main(){
    int N, K;
    cin >> N >> K;
    vector<double> X(N), Y(N), C(N);
    for(int i=0; i<N; i++) cin >> X[i] >> Y[i] >> C[i];


    auto check = [&](double T)->bool{
        vector<double> R(N);
        for(int i=0; i<N; i++) R[i] = T/C[i];
        vector<pair<double, double>> cand;
        for(int i=0; i<N; i++) cand.emplace_back(X[i], Y[i]);
        for(int i=0; i<N; i++) for(int j=i+1; j<N; j++){
            auto res = crosspoint(X[i], Y[i], R[i], X[j], Y[j], R[j]);
            for(auto& p : res) cand.push_back(p);
        }
        for(auto& p : cand){
            double x = p.first, y = p.second;
            int num = 0;
            for(int i=0; i<N; i++) if(hypot(x-X[i], y-Y[i]) <= R[i] + EPS) num++;
            if(num >= K) return true;
        }
        return false;
    };

    double ok = 1e9, ng = 0;
    for(int t=0; t<100; t++){
        double mid = (ok+ng)/2;
        (check(mid) ? ok : ng) = mid;
    }
    cout << fixed << setprecision(10) << ok << endl;
    return 0;
}