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>
#include <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <map>
#include <iomanip>
#include <stdlib.h>
#include <queue>
#include <deque>
#include <set>
#include <stack>
#include <time.h>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> Pii;
typedef pair<int, ll> Pil;
typedef pair<ll, ll> Pll;
typedef pair<ll, int> Pli;

const ll nmax = 1e9 + 7;
const ll Mod = 998244353;
const double PI = 2 * asin(1);

ld Calc_Dist (ld X1, ld Y1, ld X2, ld Y2){
  ld ans = (X1 - X2) * (X1 - X2) + (Y1 - Y2) * (Y1 - Y2);
  ans = sqrtl(ans);
  return ans;
}

int main(){
  const ld EPS = 1e-10;
  int N, K; cin >> N >> K;

  ld X[N], Y[N], C[N];
  for (int i = 0; i < N; i++){
    cin >> X[i] >> Y[i] >> C[i];
  }

  ld L = 0, R = 1e10;
  while (R - L > EPS){
    ld now = (L + R) / 2;

    vector <pair<ld, ld>> Cand;


    for (int i = 0; i < N; i++){
      Cand.push_back(make_pair(X[i], Y[i]));
      for (int j = i + 1; j < N; j++){
        ld Ri = now / C[i], Rj = now / C[j];
        ld Dist = Calc_Dist(X[i], Y[i], X[j], Y[j]);

        if (Dist < abs(Ri - Rj) - EPS) continue;
        if (Dist > Ri + Rj + EPS) continue;

        double theta = atan((Y[j] - Y[i]) / (X[j] - X[i]));
        
        if (X[j] - X[i] <= 0) theta += PI;
        else if (Y[j] - Y[i] <= 0) theta += 2 * PI;

        double phi = acos((Ri * Ri + Dist * Dist - Rj * Rj) / (2 * Ri * Dist));

        double Xnow = X[i] + Ri * cos(theta + phi);
        double Ynow = Y[i] + Ri * sin(theta + phi);
        Cand.push_back(make_pair(Xnow, Ynow));

        Xnow = X[i] + Ri * cos(theta - phi);
        Ynow = Y[i] + Ri * sin(theta - phi);
        Cand.push_back(make_pair(Xnow, Ynow));
      }
    }
    
    bool flag = false;
    for (int i = 0; i < Cand.size(); i++){
      int cnt = 0;
      ld X_now = Cand[i].first, Y_now = Cand[i].second;

      for (int j = 0; j < N; j++){
        ld Dist = Calc_Dist(X_now, Y_now, X[j], Y[j]);
        if (Dist <= now / C[j] + EPS) cnt++;
      }
      if (cnt >= K) flag = true;
    }
    
    if (flag == true) R = now;
    else L = now;
  }
  cout << setprecision(16) << (L + R) / 2 << endl;

  return 0;
}