#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <queue>

using namespace std;

int n, k;
vector<int> x, y, c;

double distance(double px, double py) {
  vector<double> dis(n);
  for (int i = 0; i < n; ++i) {
    dis[i] = c[i] * hypot(px - x[i], py - y[i]);
  }
  nth_element(dis.begin(), dis.begin() + k - 1, dis.end());
  return dis[k - 1];
}

double distance_lb(double px, double py, double r) {
  vector<double> dis(n);
  for (int i = 0; i < n; ++i) {
    dis[i] = c[i] * hypot(max(abs(px - x[i]) - r, 0.0), max(abs(py - y[i]) - r, 0.0));
  }
  nth_element(dis.begin(), dis.begin() + k - 1, dis.end());
  return dis[k - 1];
}

int main() {
  cin >> n >> k;
  x.resize(n);
  y.resize(n);
  c.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i] >> c[i];
  }
  queue<tuple<double, double, double>> que;
  double res = 1e9;
  que.emplace(0, 0, 1000);
  while (!que.empty()) {
    double px, py, r;
    tie(px, py, r) = que.front();
    que.pop();
    double lb = distance_lb(px, py, r);
    if (res - lb > 5e-7 * max(1.0, lb)) {
      res = min(res, distance(px, py));
      for (double dx : {-0.5, 0.5}) {
        for (double dy : {-0.5, 0.5}) {
          que.emplace(px + dx * r, py + dy * r, r / 2);
        }
      }
    }
  }
  cout << fixed << setprecision(15) << res << endl;
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <queue>

using namespace std;

int n, k;
vector<int> x, y, c;

double distance(double px, double py) {
  vector<double> dis(n);
  for (int i = 0; i < n; ++i) {
    dis[i] = c[i] * hypot(px - x[i], py - y[i]);
  }
  nth_element(dis.begin(), dis.begin() + k - 1, dis.end());
  return dis[k - 1];
}

double distance_lb(double px, double py, double r) {
  vector<double> dis(n);
  for (int i = 0; i < n; ++i) {
    dis[i] = c[i] * hypot(max(abs(px - x[i]) - r, 0.0), max(abs(py - y[i]) - r, 0.0));
  }
  nth_element(dis.begin(), dis.begin() + k - 1, dis.end());
  return dis[k - 1];
}

int main() {
  cin >> n >> k;
  x.resize(n);
  y.resize(n);
  c.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i] >> c[i];
  }
  queue<tuple<double, double, double>> que;
  double res = 1e9;
  que.emplace(0, 0, 1000);
  while (!que.empty()) {
    double px, py, r;
    tie(px, py, r) = que.front();
    que.pop();
    double lb = distance_lb(px, py, r);
    if (res - lb > 5e-7 * max(1.0, lb)) {
      res = min(res, distance(px, py));
      for (double dx : {-0.5, 0.5}) {
        for (double dy : {-0.5, 0.5}) {
          que.emplace(px + dx * r, py + dy * r, r / 2);
        }
      }
    }
  }
  cout << fixed << setprecision(15) << res << endl;
}

```