#include<bits/stdc++.h>

const long double EPS = 1e-10;

using namespace std;

const complex<long double> I = complex<long double>(0, 1);

int main(){
  int N, K;
  cin >> N >> K;
  
  vector<complex<long double>> meats(N);
  vector<long double> c(N);
  for(int i = 0; i < N; i++){
    long double re, im;
    cin >> re >> im >> c[i];
    meats[i] = complex<long double>(re, im);
  }
  
  long double ok = 200000, ng = 0;
  for(int k = 0; k < 100; k++){
    long double mid = (ok + ng) * 0.5;
    
    vector<complex<long double>> candidates = meats;
    for(int i = 0; i < N; i++){
      for(int j = i + 1; j < N; j++){
        long double d = abs(meats[i] - meats[j]);
        
        if(d > mid / c[i] + mid / c[j] - EPS) continue;
        if(abs(mid / c[j] - mid / c[i]) > d - EPS) continue;
        
        long double t = (mid / c[i]) * (mid / c[i]) - (mid / c[j]) * (mid / c[j]) + d * d;
        t /= d * 2;
        
        if((mid / c[i]) * (mid / c[i]) - t * t < EPS) continue;
        
        complex<long double> dir = (meats[i] - meats[j]) / d * sqrt((mid / c[i]) * (mid / c[i]) - t * t) * I;
        
        candidates.push_back(meats[i] + (meats[j] - meats[i]) / d * t + dir);
        candidates.push_back(meats[i] + (meats[j] - meats[i]) / d * t - dir);
      }
    }
    
    bool good = false;
    for(auto heat_source : candidates){
      int count = 0;
      for(int i = 0; i < N; i++){
        if(abs(heat_source - meats[i]) <= mid / c[i] + EPS){
          ++count;
        }
      }
      
      if(count >= K){
        good = true;
        break;
      }
    }
    
    (good ? ok : ng) = mid;
  }
  
  cout << fixed << setprecision(10) << ok << endl;
  
  return 0;
}
### Prompt

Create a solution in Cpp for the following 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

### Response

```cpp
#include<bits/stdc++.h>

const long double EPS = 1e-10;

using namespace std;

const complex<long double> I = complex<long double>(0, 1);

int main(){
  int N, K;
  cin >> N >> K;
  
  vector<complex<long double>> meats(N);
  vector<long double> c(N);
  for(int i = 0; i < N; i++){
    long double re, im;
    cin >> re >> im >> c[i];
    meats[i] = complex<long double>(re, im);
  }
  
  long double ok = 200000, ng = 0;
  for(int k = 0; k < 100; k++){
    long double mid = (ok + ng) * 0.5;
    
    vector<complex<long double>> candidates = meats;
    for(int i = 0; i < N; i++){
      for(int j = i + 1; j < N; j++){
        long double d = abs(meats[i] - meats[j]);
        
        if(d > mid / c[i] + mid / c[j] - EPS) continue;
        if(abs(mid / c[j] - mid / c[i]) > d - EPS) continue;
        
        long double t = (mid / c[i]) * (mid / c[i]) - (mid / c[j]) * (mid / c[j]) + d * d;
        t /= d * 2;
        
        if((mid / c[i]) * (mid / c[i]) - t * t < EPS) continue;
        
        complex<long double> dir = (meats[i] - meats[j]) / d * sqrt((mid / c[i]) * (mid / c[i]) - t * t) * I;
        
        candidates.push_back(meats[i] + (meats[j] - meats[i]) / d * t + dir);
        candidates.push_back(meats[i] + (meats[j] - meats[i]) / d * t - dir);
      }
    }
    
    bool good = false;
    for(auto heat_source : candidates){
      int count = 0;
      for(int i = 0; i < N; i++){
        if(abs(heat_source - meats[i]) <= mid / c[i] + EPS){
          ++count;
        }
      }
      
      if(count >= K){
        good = true;
        break;
      }
    }
    
    (good ? ok : ng) = mid;
  }
  
  cout << fixed << setprecision(10) << ok << endl;
  
  return 0;
}
```