Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 3
-1 0 3
0 0 3
1 0 2
1 1 40

Sample Output 1

2.4

If we put the heat source at \left(-0.2, 0\right), the 1-st, 2-nd, and 3-rd pieces of meat will be ready to eat within 2.4 seconds. This is the optimal place to put the heat source.

Sample Input 2

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

Sample Output 2

7411.2252

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T, class U> using Pa = pair<T, U>;
template <class T> using vec = vector<T>;
template <class T> using vvec = vector<vec<T>>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,K;
    cin >> N >> K;
    vec<double> X(N),Y(N),C(N);
    for(int i=0;i<N;i++) cin >> X[i] >> Y[i] >> C[i];
    double ans = 1e18;
    
    auto calc = [&](double x,double y){
        vec<double> S(N);
        for(int i=0;i<N;i++){
            double dx = X[i]-x,dy = Y[i]-y;
            S[i] = C[i]*sqrt(dx*dx+dy*dy);
        }
        sort(S.begin(),S.end());
        S.resize(K);
        double res = *max_element(S.begin(),S.end());
        ans = min(ans,res);
        return res;
    };

    auto search_y = [&](double y){
        double l = -100000,r = 100000;
        for(int i=0;i<100;i++){
            double ml = (2*l+r)/3,mr = (l+2*r)/3;
            if(calc(ml,y)<calc(mr,y)) r = mr;
            else l = ml;
        }
        return calc(l,y);
    };

    double l = -100000,r = 100000;
    for(int i=0;i<100;i++){
        double ml = (2*l+r)/3,mr = (l+2*r)/3;
        if(search_y(ml)<search_y(mr)) r = mr;
        else l = ml;
    }
    cout << fixed << setprecision(10) << ans << "\n";
}