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 <stdio.h>
#include <queue>
#include <math.h>
#include <utility>
#define delta 1e-7
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, l, m;
  scanf("%lld%lld", &n, &k);
  ll x[n], y[n], c[n];
  for(i = 0; i < n; ++i) scanf("%lld%lld%lld", &x[i], &y[i], &c[i]);
  double he = 0.0, ta = 500000.0, r;
  for(m = 0; m < 50; ++m) {
    r = (he + ta) / 2.0;
    queue< pair<double, double> > q;
    for(i = 0; i < n; ++i) q.push(make_pair((double)x[i], (double)y[i]));
    for(i = 0; i < n; ++i) for(j = i + 1; j < n; ++j) 
      if((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) <= (r / c[i] + r / c[j]) * (r / c[i] + r / c[j]) + delta && (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) >= (r / c[i] - r / c[j]) * (r / c[i] - r / c[j]) + delta) {
      double a = x[j] - x[i], b = y[j] - y[i], r1 = r / c[i], r2 = r / c[j];
      if(y[j] != y[i]) {
        double t = -a / b, s = (a * a + b * b + r1 * r1 - r2 * r2) / (2 * b);
        double d = sqrt(t * t * s * s - (1 + t * t) * (s * s - r1 * r1)), xt = (-t * s + d) / (1 + t * t);
        q.push(make_pair(xt + x[i], t * xt + s + y[i]));
        xt = (-t * s - d) / (1 + t * t);
        q.push(make_pair(xt + x[i], t * xt + s + y[i]));
      } else {
        double xt = (a * a + r1 * r1 - r2 * r2) / (2 * a);
        double yt = sqrt(r1 * r1 - xt * xt);
        q.push(make_pair(xt + x[i], yt + y[i])), q.push(make_pair(xt + x[i], -yt + y[i]));
      }
    }
    while(q.size()) {
      ll cnt = 0;
      for(i = 0; i < n; ++i) {
        if((x[i] - q.front().first) * (x[i] - q.front().first) + (y[i] - q.front().second) * (y[i] - q.front().second) <= r / c[i] * r / c[i] + delta) cnt++;
      }
      if(cnt >= k) {
        ta = r;
        break;
      }
      q.pop();
    }
    if(!q.size()) he = r;
  }
  printf("%lf", r);
  return 0;
}