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 lint = long long int;
using P = complex<double>;

// [left, right) の乱数生成
template<class T> T randrange(T left, T right) {
    static mt19937 mt{ std::random_device{}() };
    uniform_real_distribution<double> rnd(left, right);
    return (is_integral<T>::value ? floor(rnd(mt)) : rnd(mt));
}

int main() {
    lint n, k;
    cin >> n >> k;

    vector<pair<P, lint>> p(n);
    for(lint i=0; i<n; i++) {
        lint x, y, c;
        cin >> x >> y >> c;
        p[i] = make_pair(P(x, y), c);
    }

    auto dist = [&](P c) {
        vector<double> cand;
        for(auto& e : p) {
            P point = e.first;
            lint cost = e.second;

            cand.push_back(cost * abs(point - c));
        }
        sort(cand.begin(), cand.end());
        double res = cand[k-1];
        return res;
    };

    P c = {0, 0};
    double r = 1.0;
    for(lint _=0; _<4e5; _++) {
        P next = c + P(randrange(-r, r), randrange(-r, r));
        if(dist(next) < dist(c))  c = next;
        r *= 0.99999;
    }

    cout << setprecision(18) << fixed;
    cout << dist(c) << endl;
}