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:

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

constexpr double eps = 1e-9;
constexpr double inf = 1e6;
constexpr bool eq(double x, double y) {
  return fabs(x - y) < eps;
}

struct P {
  double x, y;
  P(double x = 0, double y = 0): x(x), y(y) { }
  P rot() const { return P(-y, x); }
  P operator + (P p) const { return P(x + p.x, y + p.y); }
  P operator - (P p) const { return P(x - p.x, y - p.y); }
  P operator * (double k) const { return P(x * k, y * k); }
  double nm() const { return sqrt(x * x + y * y); }
};

struct C {
  P o; double r;
  C(P o = P(), double r = 0): o(o), r(r) { }
  vector<P> sc(C c) {
    P p = c.o - o;
    double l = p.nm();
    if (eq(l, 0)) {
      return {};
    }
    if (eq(l + r + c.r, max({ l, r, c.r }) * 2.0)) {
      return { o + p * (r / l) };
    }
    if (l + r + c.r < max({ l, r, c.r }) * 2.0) {
      return {};
    }
    double x = -(c.r * c.r - l * l - r * r) / (2.0 * l);
    double y = sqrt(r * r - x * x);
    P md = o + p * (x / l);
    p = p.rot();
    return { md + p * (y / l), md - p * (y / l) };
  }
  bool con(P p) {
    return (p - o).nm() < r + eps;
  }
};

int main() {
  int N, K;
  scanf("%d %d", &N, &K);
  if (K == 1) {
    puts("0");
    return 0;
  }
  vector<C> cs(N);
  vector<double> coeff(N);
  for (int i = 0; i < N; ++i) {
    scanf("%lf %lf %lf", &cs[i].o.x, &cs[i].o.y, &coeff[i]);
  }
  double ok = inf, ng = 0;
  for (int step = 0; step < 60; ++step) {
    double md = (ok + ng) / 2;
    vector<P> ps;
    for (int i = 0; i < N; ++i) {
      cs[i].r = md / coeff[i];
      ps.push_back(cs[i].o);
      for (int j = 0; j < i; ++j) {
        for (auto p: cs[i].sc(cs[j])) {
          ps.push_back(p);
        }
      }
    }
    ([&] {
      for (auto p: ps) {
        int cnt = 0;
        for (auto c: cs) {
          if (c.con(p)) {
            ++cnt;
          }
        }
        if (cnt >= K) {
          return true;
        }
      }
      return false;
    }() ? ok : ng) = md;
  }
  printf("%.10lf\n", ok);
  return 0;
}