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; inline void fileio(const char* in, const char* out) { freopen(in, "r", stdin); freopen(out, "w", stdout); }
#define TD typedef
#define forx(i,n) for(int i = 0; i < (n); ++i)
#define forxr(i,n) for(int i = (n)-1; i >= 0; --i)
TD long long ll; TD long double ld; TD pair<int,int> pii; TD pair<ll,ll> pll; TD vector<int> VI; TD vector<bool> VB; TD vector<ll> VL;
const char nl = '\n'; const int INF = 0x3f3f3f3f; const ll LINF = 0x3f3f3f3f3f3f3f3fll; const ld EPS = 1e-9, PI = acos(-1);
#define ff first
#define ss second
#define all(c) c.begin(),c.end()

using pt = complex<ld>;
const int N = 60;
int n, k;
ld x[N], y[N], c[N];

vector<pt> cxc(pt c1, ld r1, pt c2, ld r2) {
  ld dq = norm(c1-c2), rq=r1*r1-r2*r2;
  pt c = (c1+c2)+(c2-c1)*rq/dq; c*=.5;
  ld dt=2*dq*(r1*r1+r2*r2)-dq*dq-rq*rq;
  if (dt < -EPS) return {};
  if (dt < EPS) return {c};
  dt=sqrt(dt)/2/dq;
  pt d = (c2-c1)*pt(0,1)*dt;
  return {c+d, c-d};
}

bool poss(ld t) {
  for (int i = 0; i < n; ++i) {
    for (int j = i+1; j < n; ++j) {
      auto pts = cxc({x[i], y[i]}, t/c[i], {x[j], y[j]}, t/c[j]);
      for (pt p : pts) {
        int cnt = 0;
        for (int l = 0; l < n; ++l) {
          pt o = {x[l], y[l]};
          ld d = abs(o-p);
          if (d - EPS < t/c[l]) {
            cnt++;
            if (cnt >= k) return true;
          }
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    pt p(x[i], y[i]);
    int cnt = 0;
    for (int l = 0; l < n; ++l) {
      pt o = {x[l], y[l]};
      ld d = abs(o-p);
      if (d - EPS < t/c[l]) {
        cnt++;
        if (cnt >= k) return true;
      }
    }
  }
  return false;
}

int main() { // Emily <3
  atexit([](){ cerr << "Time: " << (ld)clock() / CLOCKS_PER_SEC << nl; });
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  cout << fixed << setprecision(12);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i] >> c[i];
  }
  ld lo = 0, hi = 1e10;
  while (abs(hi - lo) > EPS) {
    ld mid = (lo+hi)/2;
    if (poss(mid)) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  cout << hi << nl;

  return 0;
}