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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const double eps = 1e-9;
bool equal(double a, double b) { return abs(a-b) < eps;}

struct V {
  double x, y;
  V(double x=0, double y=0) : x(x), y(y) {}
  V operator+(const V& p) const { return V(x+p.x, y+p.y);}
  V operator-(const V& p) const { return V(x-p.x, y-p.y);}
  V operator*(double a) const { return V(x*a, y*a);}
  double sqrNorm() const { return x*x+y*y;}
  double norm() const { return sqrt(sqrNorm());}
  V rotate90() const { return V(y, -x);}
};

struct Circle {
  V o; double r;
  Circle(V o=V(), double r=0) : o(o), r(r) {}
  vector<V> xp(const Circle& c) {
    V v = c.o-o;
    double l = v.norm();
    if (equal(l, 0)) return {};
    if (equal(l+r+c.r, max({l,r,c.r})*2)) {
      return {o + v*(r/l)};
    }
    if (l+r+c.r < max({l,r,c.r})*2) return {};
    double x = -(c.r*c.r - l*l - r*r) / (2*l);
    double y = sqrt(r*r - x*x);
    V mid = o + v*(x/l);
    v = v.rotate90();
    return {mid + v*(y/l), mid - v*(y/l)};
  }
  bool isInside(const V& p) {
    return (p-o).norm() < r+eps;
  }
};

int main() {
  int n, k;
  cin >> n >> k;
  if (k == 1) {
    cout << 0 << endl;
    return 0;
  }
  vector<Circle> c(n);
  rep(i,n) {
    cin >> c[i].o.x >> c[i].o.y >> c[i].r;
  }
  double l = 0, r = 400000;
  rep(ti,100) {
    double x = (l+r)/2;
    vector<V> ps;
    rep(i,n) ps.push_back(c[i].o);
    rep(i,n)rep(j,i) {
      Circle a = c[i], b = c[j];
      a.r = x/a.r;
      b.r = x/b.r;
      auto tmp = a.xp(b);
      ps.insert(ps.end(), tmp.begin(), tmp.end());
    }
    bool ok = false;
    for (V p : ps) {
      int cnt = 0;
      rep(ci,n) {
        Circle nc = c[ci];
        nc.r = x/nc.r;
        if (nc.isInside(p)) cnt++;
      }
      if (cnt >= k) ok = true;
    }
    if (ok) r = x; else l = x;
  }
  printf("%.10f\n", r);
  return 0;
}