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;

double inf = 1e100;
int n, k;
double x[60];
double y[60];
double c[60];
double dist(double ax, double ay, double bx, double by) {
  double dx = bx - ax;
  double dy = by - ay;
  return hypot(dx, dy);
}

bool ok(double cx, double cy, double t) {
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    double d = dist(cx, cy, x[i], y[i]);
    if (t >= c[i] * d - 1e-10) ++cnt;
  }
  return cnt >= k;
}

pair<double, double> inter(
  double ax, double ay, double a,
  double bx, double by, double b) {
  double d = dist(ax, ay, bx, by);
  if (d >= a + b) return make_pair(-inf, -inf);
  if (d <= abs(b - a)) return make_pair(-inf, -inf);
  double t = (a + b + d) / 2;
  double s = sqrt(t * (t - a) * (t - b) * (t - d));
  double h = 2 * s / d;
  double x = sqrt(a * a - h * h);
  double y = sqrt(b * b - h * h);
  double k = y > d && y > x ? -x : x;
  
  double dx = bx - ax;
  double dy = by - ay;
  double ux = dx / d;
  double uy = dy / d;
  double mx = ax + k / d * dx;
  double my = ay + k / d * dy;
  return make_pair(mx - h * uy, my + h * ux);
}

bool sub(double t) {
  for (int i = 0; i < n; ++i) {
    if (ok(x[i], y[i], t)) return true;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      auto p = inter(x[i], y[i], t / c[i], x[j], y[j], t / c[j]);
      if (p.first == inf) continue;
      if (ok(p.first, p.second, t)) return true;
    }
  }
  return false;
}

int main() {
  cin >> n >> k;
  if (k == 1) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i] >> c[i];
  }
  
  double lo = 0;
  double hi = 1e10;
  for (int z = 0; z < 100; ++z) {
    double mid = (lo + hi) / 2;
    if (sub(mid)) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  cout << setprecision(16) << fixed << lo << endl;
}