#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<iomanip>
#include<cmath>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ld,ld> P;
ld cx[65]={0},cy[65]={0},kd[65]={0};
ll n,k;
const ld EPS=1e-14;
bool judge(ld T){
  ld cd[65];
  for(int i=0;i<n;i++) cd[i]=T/kd[i];
  vector<P> points;
  for(int i=0;i<n;i++){
    for(int j=i+1;j<n;j++){
      ld x1=cx[i],x2=cx[j],y1=cy[i],y2=cy[j];
      ld dist=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
      ld d1=cd[i],d2=cd[j];
      if(d1+d2<dist+EPS) continue;
      if(dist<max(d1-d2,d2-d1)+EPS) continue;

      ld cos=(d1*d1+dist*dist-d2*d2)/(2*d1*dist);
      ld sin1=sqrt(abs(1-cos*cos));
      ld sin2=-sin1;
      ld dx=(x2-x1)*d1/dist,dy=(y2-y1)*d1/dist;
      ld vx1=dx*cos-dy*sin1,vx2=dx*cos-dy*sin2;
      ld vy1=dx*sin1+dy*cos,vy2=dx*sin2+dy*cos;

      points.push_back(P(x1+vx1,y1+vy1));
      points.push_back(P(x1+vx2,y1+vy2));
    }
    points.push_back(P(cx[i],cy[i]));
  }
  for(int i=0;i<points.size();i++){
    int cnt=0;
    ld px=points[i].first,py=points[i].second;
    for(int j=0;j<n;j++){
      ld ox=cx[j],oy=cy[j],d=cd[j];
      ld dist=sqrt((ox-px)*(ox-px)+(oy-py)*(oy-py));
      if(dist<=d+EPS) cnt++;
    }
    if(cnt>=k) return true;
  }
  return false;
}
int main(){
  cin>>n>>k;
  for(int i=0;i<n;i++) cin>>cx[i]>>cy[i]>>kd[i];
  ld lb=0,rb=1e6;
  for(int i=0;i<500;i++){
    ld mid=(lb+rb)/2;
    if(judge(mid)) rb=mid;
    else lb=mid;
  }
  cout<<fixed;
  cout<<setprecision(10)<<rb<<endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<iomanip>
#include<cmath>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ld,ld> P;
ld cx[65]={0},cy[65]={0},kd[65]={0};
ll n,k;
const ld EPS=1e-14;
bool judge(ld T){
  ld cd[65];
  for(int i=0;i<n;i++) cd[i]=T/kd[i];
  vector<P> points;
  for(int i=0;i<n;i++){
    for(int j=i+1;j<n;j++){
      ld x1=cx[i],x2=cx[j],y1=cy[i],y2=cy[j];
      ld dist=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
      ld d1=cd[i],d2=cd[j];
      if(d1+d2<dist+EPS) continue;
      if(dist<max(d1-d2,d2-d1)+EPS) continue;

      ld cos=(d1*d1+dist*dist-d2*d2)/(2*d1*dist);
      ld sin1=sqrt(abs(1-cos*cos));
      ld sin2=-sin1;
      ld dx=(x2-x1)*d1/dist,dy=(y2-y1)*d1/dist;
      ld vx1=dx*cos-dy*sin1,vx2=dx*cos-dy*sin2;
      ld vy1=dx*sin1+dy*cos,vy2=dx*sin2+dy*cos;

      points.push_back(P(x1+vx1,y1+vy1));
      points.push_back(P(x1+vx2,y1+vy2));
    }
    points.push_back(P(cx[i],cy[i]));
  }
  for(int i=0;i<points.size();i++){
    int cnt=0;
    ld px=points[i].first,py=points[i].second;
    for(int j=0;j<n;j++){
      ld ox=cx[j],oy=cy[j],d=cd[j];
      ld dist=sqrt((ox-px)*(ox-px)+(oy-py)*(oy-py));
      if(dist<=d+EPS) cnt++;
    }
    if(cnt>=k) return true;
  }
  return false;
}
int main(){
  cin>>n>>k;
  for(int i=0;i<n;i++) cin>>cx[i]>>cy[i]>>kd[i];
  ld lb=0,rb=1e6;
  for(int i=0;i<500;i++){
    ld mid=(lb+rb)/2;
    if(judge(mid)) rb=mid;
    else lb=mid;
  }
  cout<<fixed;
  cout<<setprecision(10)<<rb<<endl;
}

```