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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;
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*(const V& p) const { return V(x*p.x - y*p.y, x*p.y + y*p.x);}
  double sqrNorm() const { return x*x+y*y;}
  double norm() const { return sqrt(sqrNorm());}
};

const double eps = 1e-9;
struct Circle {
  V o; double r;
  Circle(V o=V(), double r=0) : o(o), r(r) {}
  vector<V> xv(const Circle& c) {
    V v = c.o-o;
    double d = v.norm();
    if ( abs(d) < eps ) {
      return {};
    }
    double rc = (d*d + r*r - c.r*c.r)/(2*d);
    if ( r < rc ) {
      return {};
    }
    double rs = sqrt(r*r - rc*rc);
    v.x /= d;
    v.y /= d;
    if (rs < eps) {
      return { o + v * V(rc, 0 ) } ;
    }
    return { o + v * V(rc,rs), o + v * V(rc,-rs) };
  }
  bool isInside(const V& p) {
    return (p-o).norm() < r+eps;
  }
};

int main() {
  int n,k;
  cin >> n >> k;
  vector<Circle> c(n);
  rep(i,n){
    cin >> c[i].o.x >> c[i].o.y >> c[i].r;
  }
  if ( k == 0 ){
    cout << 0 << endl;
    return 0;
  }
  double r = 0, l = 200000;
  rep(xxx,100){
    double x = (r+l)/2;

    vector<V> p;
    rep(i,n) p.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 v = a.xv(b);
      p.insert(p.end(),v.begin(),v.end());
    }
    
    int cnt = 0;
    for ( auto pi : p ){
      cnt = 0;
      rep(i,n){
        auto ci = Circle( c[i].o,c[i].r);
        ci.r = x/ci.r;
        if (ci.isInside(pi)) {
          ++cnt;
        }
      }
      if (cnt >= k) {
        break;
      }
    }
    if ( cnt >= k ) l = x;
    else r = x;
  }
  printf("%.10f\n",r);  
}