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:
  #define _USE_MATH_DEFINES
  #include <bits/stdc++.h>
  using namespace std;

  struct Circle {
    double x, y, r;
    Circle (double xx, double yy, double rr) : x(xx), y(yy), r(rr) {}
  };

  vector<pair<double, double>> inter (Circle A, Circle B) {
  double d = hypot(B.x - A.x, B.y - A.y);
  if (d <= A.r + B.r + 1e-7 && d + 1e-7 >= abs(B.r - A.r)) {

      double ex = (B.x - A.x) / d;
      double ey = (B.y - A.y) / d;

      double x = (A.r * A.r - B.r * B.r + d * d) / (2 * d);
      double y = sqrt(A.r * A.r - x * x);
      vector<pair<double, double>> res;
      pair<double, double> P1 = {A.x + x * ex - y * ey, A.y + x * ey + y * ex};

      pair<double, double> P2 = {A.x + x * ex + y * ey, A.y + x * ey - y * ex};
      return {P1, P2};
  } else {
    return {{0, 0}};
  }
  }

  signed main() { 
    ios::sync_with_stdio(false); cin.tie(0);
    //f
    int n, k;
    cin >> n >> k;
    vector<vector<double>> in;
    for (int i = 0; i < n; i++) {
      double x, y, c;
      cin >> x >> y >> c;
      in.push_back({x, y, c});
    }
    if (k == 1) {
      cout << 0 << endl;
      return 0;
    }
    auto judge = [&] (double time) -> bool {
      vector<Circle> v;
      for (int i = 0; i < n; i++) {
        v.push_back(Circle(in[i][0], in[i][1], time / in[i][2]));
      }
      for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) {
        auto res = inter(v[i], v[j]);
        for (auto& p : res) {
          int cnt = 0;
          for (int u = 0; u < n; u++) {
            if (hypot(v[u].x - p.first, v[u].y - p.second) * in[u][2] <= time + 1e-7) {
              cnt++;
            }
          }
          if (cnt >= k) return true;
        }
      }
      for (int i = 0; i < n; i++) {
        int cnt = 0;
        for (int j = 0; j < n; j++) {
          if (hypot(v[i].x - v[j].x, v[i].y - v[j].y) * in[j][2] <= time + 1e-7) {
            cnt++;
          }
        }
        if (cnt >= k) return true;
      }
      return false;
    };
    //cerr << judge(3) << endl;
    //return 0;
    double lo = 0, hi = 1e9;
    for (int i = 0; i < 200; i++) {
      double mid = (lo + hi) / 2;
      //cerr << mid << " " << judge(mid) << endl;
      if (judge(mid)) hi = mid;
      else lo = mid;
    }
    cout << setprecision(20) << fixed << hi << endl;
    return 0;
  }