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>
#include <cctype>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//template<typename key, typename val>
//using ordered_tree =
//  tree<key, val, std::less<>, rb_tree_tag, tree_order_statistics_node_update>;
using namespace std;
typedef long long int64;
typedef unsigned long long uint64;
typedef complex<double> comp;
const double pi = 3.14159265358979323846;
const int inf = (int)1e+9 + 2;
const int64 inf64 = (int64)1e+18 + 2;
const double dinf = 1e+20;
const int mod = 1'000'000'007;//*/998244353;
const int base = 2187;
const double eps = 1e-9;
const int N = 300'000;
const int LOGN = 19;

int n, m, k;

int x[N], y[N], c[N];
double d[N];

int iter = 900;

double f(double xx, double yy) {
  for (int i = 0; i < n; ++i) {
    d[i] = c[i] * c[i] * ((x[i] - xx) * (x[i] - xx) + (y[i] - yy) * (y[i] - yy));
  }
  sort(begin(d), begin(d) + n);
  return d[k - 1];
}

double g(double xx) {
  double l = -1000, r = 1000;
  for (int i = 0; i < iter; ++i) {
    double mid = (r - l) / 3;
    if (f(xx, l + mid) < f(xx, l + mid + mid)) {
      r = l + mid + mid;
    } else {
      l = l + mid;
    }
  }
  return f(xx, l);
}

void solve(int) {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i] >> c[i];
  }
  double l = -1000, r = 1000;
  for (int i = 0; i < iter; ++i) {
    double mid = (r - l) / 3;
    if (g(l + mid) < g(l + mid + mid)) {
      r = l + mid + mid;
    } else {
      l = l + mid;
    }
  }
  cout.setf(ios::fixed);
  cout.precision(11);
  cout << sqrt(g(l)) << '\n';
}

void precalc() {
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  precalc();
  int test = 1;
  //cin >> test;
  //auto start = chrono::high_resolution_clock::now();
  for (int i = 1; i <= test; ++i) {
    solve(i);
  }
  //cerr << chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - start).count() << '\n';
  return 0;
}