Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 3
-1 0 3
0 0 3
1 0 2
1 1 40

Sample Output 1

2.4

If we put the heat source at \left(-0.2, 0\right), the 1-st, 2-nd, and 3-rd pieces of meat will be ready to eat within 2.4 seconds. This is the optimal place to put the heat source.

Sample Input 2

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

Sample Output 2

7411.2252

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cassert>
#include <cmath>
#include <iomanip>
#include <vector>

using namespace std;

typedef long double ld;
typedef pair<ld, ld> pt;
#define x first
#define y second

pt operator+ (pt p, pt q) {
  p.x += q.x;
  p.y += q.y;
  return p;
}

pt operator- (pt p) {
  p.x *= -1;
  p.y *= -1;
  return p;
}

pt operator- (pt p, pt q) {
  return p + (-q);
}

ld dot (pt p, pt q) {
  return p.x * q.x + p.y * q.y;
}

ld sq (pt p) {
  return dot(p, p);
}

pt operator* (pt p, double k) {
  p.x *= k;
  p.y *= k;
  return p;
}

pt operator/ (pt p, double k) {
  p.x /= k;
  p.y /= k;
  return p;
}

pt perp (pt p) {
  return make_pair(-p.y, p.x);
}

const ld SGEPS = 1e-8;
int sgn (ld x) {
  if (x < 0) return -1;
  if (x > 0) return 1;
  return 0;
}

int circleCircle(pt o1, ld r1s, pt o2, ld r2s, pair<pt,pt> &out) {
  pt d=o2-o1; ld d2=sq(d);
  ld pd = (d2 + r1s - r2s)/2; // = |O_1P| * d
  ld h2 = r1s - pd*pd/d2; // = h^2
  if (h2 >= 0) {
    pt p = o1 + d*pd/d2, h = perp(d)*sqrt(h2/d2);
    out = {p-h, p+h};
  } else if (h2 > -1e-15) {
    pt p = o1 + d * (pd / d2);
    out = {p, p};
    return 1;
  }
  return 1 + sgn(h2);
}

const ld EPS = 1e-10;
int inCircle (pt o, ld r1s, pt p) {
  return sq(p - o) < r1s + EPS;
}

const int MAX_N = 65;

int n;
pt pts [MAX_N];
ld speed [MAX_N];
ld rsq [MAX_N];

int countIn (pt p) {
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (inCircle(pts[i], rsq[i], p)) {
      ans++;
    }
  }
  return ans;
}

ld sq (ld x) {
  return x * x;
}

int k;
bool canGet (double t) {
  for (int i = 0; i < n; i++) {
    rsq[i] = sq(t / speed[i]);
  }

  for (int i = 0; i < n; i++) {
    if (countIn(pts[i]) >= k) return true;
    for (int j = i + 1; j < n; j++) {
      pair<pt, pt> out;
      int c = circleCircle(pts[i], rsq[i], pts[j], rsq[j], out);
      if (c >= 1) {
        if (countIn(out.first) >= k) return true;
      }
      if (c >= 2) {
        if (countIn(out.second) >= k) return true;
      }
    }
  }
  return false;
}

int main () {
  ios::sync_with_stdio(false);

  cin >> n >> k;

  for (int i = 0; i < n; i++) {
    cin >> pts[i].x >> pts[i].y >> speed[i];
  }
  
  ld ans = 0;
  for (ld step = 1e9; step > 1e-8; step /= 2) {
    if (!canGet(ans + step)) {
      ans += step;
    }
  }

  cout << fixed << setprecision(12) << ans << endl;
}