#include <bits/stdc++.h>
using namespace std;
const int N=65;
const double eps=1e-8;
int n,k;
double x[N],y[N],c[N],r[N];
double dis(pair<double,double> a,pair<double,double> b) {
    return sqrt((a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second));
}
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(S==0)return {};
    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) {
    for(int i=0;i<n;i++) r[i]=t/c[i];
    vector<pair<double,double> > vec;
    for(int i=0;i<n;i++) {
    	vec.push_back({x[i],y[i]});
        for(int j=i+1;j<n;j++) {
            auto v=cal(x[i],y[i],r[i],x[j],y[j],r[j]);
            for(auto it:v) vec.push_back(it);
        }
    }
    for(auto it:vec) {
        int cnt=0;
        for(int i=0;i<n;i++) {
            if(r[i]+eps>=dis(it,{x[i],y[i]})) cnt++;
        }
        if(cnt>=k) return true;
    }
    return false;
}
int main() {
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++) {
        scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);
    }
    double L=0,R=1000010;
    for(int i=0;i<100;i++) {
        double mid=(L+R)/2;
        if(check(mid)) R=mid;
        else L=mid;
    }
    printf("%.6lf\n",R);
    return 0;
}
### 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;
const int N=65;
const double eps=1e-8;
int n,k;
double x[N],y[N],c[N],r[N];
double dis(pair<double,double> a,pair<double,double> b) {
    return sqrt((a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second));
}
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(S==0)return {};
    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) {
    for(int i=0;i<n;i++) r[i]=t/c[i];
    vector<pair<double,double> > vec;
    for(int i=0;i<n;i++) {
    	vec.push_back({x[i],y[i]});
        for(int j=i+1;j<n;j++) {
            auto v=cal(x[i],y[i],r[i],x[j],y[j],r[j]);
            for(auto it:v) vec.push_back(it);
        }
    }
    for(auto it:vec) {
        int cnt=0;
        for(int i=0;i<n;i++) {
            if(r[i]+eps>=dis(it,{x[i],y[i]})) cnt++;
        }
        if(cnt>=k) return true;
    }
    return false;
}
int main() {
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++) {
        scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);
    }
    double L=0,R=1000010;
    for(int i=0;i<100;i++) {
        double mid=(L+R)/2;
        if(check(mid)) R=mid;
        else L=mid;
    }
    printf("%.6lf\n",R);
    return 0;
}
```