#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
const int MAXN = 5e5+5;
const int MOD= 998244353;
int n,k;
struct node{
    double x,y,c,r;
}a[MAXN];
vector<pair<double,double> >v;
vector<pair<double,double> > cal(double x1,double y1,double r1,double x2,double y2,double r2) {//两圆交点
    x1-=x2,y1-=y2;
    double S=x1*x1+y1*y1,a=(S+r2*r2-r1*r1)/2,D=S*r2*r2-a*a;
    if(D<0) return {};
    double A1=a*x1,B1=y1*sqrt(D);
    double A2=a*y1,B2=x1*sqrt(D);
    return {{(A1+B1)/S+x2,(A2-B2)/S+y2},{(A1-B1)/S+x2,(A2+B2)/S+y2}};
}
bool check(double t){
    rep(i,1,n)a[i].r=t/a[i].c;
    v.clear();
    rep(i,1,n){
        v.push_back(make_pair(a[i].x,a[i].y));
        rep(j,i+1,n){
            auto p=cal(a[i].x,a[i].y,a[i].r,a[j].x,a[j].y,a[j].r);
            for(auto it:p){
                v.push_back(it);
            }
        }
    }
    int len=v.size();
    rep(i,0,len-1){
        pair<double,double>p=v[i];
        int num=0;
        rep(i,1,n){
            if((a[i].x-p.first)*(a[i].x-p.first)+(a[i].y-p.second)*(a[i].y-p.second)<=a[i].r*a[i].r||
               fabs((a[i].x-p.first)*(a[i].x-p.first)+(a[i].y-p.second)*(a[i].y-p.second)-a[i].r*a[i].r)<=1e-8)num++;
            if(num==k)return 1;
        }
    }return 0;
}
int main(){
    while(cin>>n>>k){
        rep(i,1,n)cin>>a[i].x>>a[i].y>>a[i].c;
        double l=0,r=1e9;
        rep(i,1,100){
            double mid=(l+r)/2;
            if(check(mid))r=mid;
            else l=mid;
        }
        printf("%.10lf\n",r);
    }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;

#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
const int MAXN = 5e5+5;
const int MOD= 998244353;
int n,k;
struct node{
    double x,y,c,r;
}a[MAXN];
vector<pair<double,double> >v;
vector<pair<double,double> > cal(double x1,double y1,double r1,double x2,double y2,double r2) {//两圆交点
    x1-=x2,y1-=y2;
    double S=x1*x1+y1*y1,a=(S+r2*r2-r1*r1)/2,D=S*r2*r2-a*a;
    if(D<0) return {};
    double A1=a*x1,B1=y1*sqrt(D);
    double A2=a*y1,B2=x1*sqrt(D);
    return {{(A1+B1)/S+x2,(A2-B2)/S+y2},{(A1-B1)/S+x2,(A2+B2)/S+y2}};
}
bool check(double t){
    rep(i,1,n)a[i].r=t/a[i].c;
    v.clear();
    rep(i,1,n){
        v.push_back(make_pair(a[i].x,a[i].y));
        rep(j,i+1,n){
            auto p=cal(a[i].x,a[i].y,a[i].r,a[j].x,a[j].y,a[j].r);
            for(auto it:p){
                v.push_back(it);
            }
        }
    }
    int len=v.size();
    rep(i,0,len-1){
        pair<double,double>p=v[i];
        int num=0;
        rep(i,1,n){
            if((a[i].x-p.first)*(a[i].x-p.first)+(a[i].y-p.second)*(a[i].y-p.second)<=a[i].r*a[i].r||
               fabs((a[i].x-p.first)*(a[i].x-p.first)+(a[i].y-p.second)*(a[i].y-p.second)-a[i].r*a[i].r)<=1e-8)num++;
            if(num==k)return 1;
        }
    }return 0;
}
int main(){
    while(cin>>n>>k){
        rep(i,1,n)cin>>a[i].x>>a[i].y>>a[i].c;
        double l=0,r=1e9;
        rep(i,1,100){
            double mid=(l+r)/2;
            if(check(mid))r=mid;
            else l=mid;
        }
        printf("%.10lf\n",r);
    }
}

```