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>
using namespace std;
typedef long long ll;
#define For(i,n,k) for(ll i=(n);i<(k);i++)
#define ALL(a)  (a).begin(),(a).end()
double ans=1e9;
double eps=1e-9;
double dist(double x,double y,double a,double b){
    return sqrt((x-a)*(x-a)+(y-b)*(y-b));
}
double dist2(double x,double y,double a,double b){
    return ((x-a)*(x-a)+(y-b)*(y-b));
}
struct Niku{
    double x;
    double y;
    double c;
};
void Main(){
    ll n;cin>>n;
    ll k;cin>>k;
    vector<Niku> niku(n);
    For(i,0,n){
        cin>>niku[i].x>>niku[i].y>>niku[i].c;
    }
    if(k==1){
        ans=0;
    }
    else{
        double ok=1e7,ng=0;
        while(ok-ng>1e-10){
            double mid=(ok+ng)/2;
            bool flag=false;
            For(i,0,n){
                ll cnt=0;
                For(j,0,n){
                    if(dist(niku[i].x,niku[i].y,niku[j].x,niku[j].y)<=mid/niku[j].c+eps) cnt++;
                }
                if(cnt>=k){
                    flag=true;
                    goto next;
                }
                For(j,i+1,n){
                    double r1=mid/niku[i].c;
                    double r2=mid/niku[j].c;
                    double D=dist(niku[i].x,niku[i].y,niku[j].x,niku[j].y);
                    if(D>r1+r2||r1>D+r2||r2>D+r1) continue;
                    double x1=(niku[j].x-niku[i].x),y1=(niku[j].y-niku[i].y);
                    double x2y2=x1*x1+y1*y1;
                    double a=(x2y2+r1*r1-r2*r2)/2/D;
                    double h=sqrt(r1*r1-a*a);
                    For(t,0,2){
                        double X=niku[i].x+a*x1/D,Y=niku[i].y+a*y1/D;
                        if(t==0){
                            X+=h*y1/D;
                            Y-=h*x1/D;
                        }
                        else{
                            X-=h*y1/D;
                            Y+=h*x1/D;
                        }
                        cnt=0;
                        For(i2,0,n){
                            if(dist(niku[i2].x,niku[i2].y,X,Y)<=mid/niku[i2].c+eps){
                                cnt++;
                            }
                        }
                        if(cnt>=k){
                            flag=true;
                            goto next;
                        }
                    }
                }
            }
            next:;
            if(flag) ok=mid;
            else ng=mid;
        }
        ans=ok;
    }
    cout<<fixed<<setprecision(10)<<ans<<endl;
}
int main(){
    Main();
    return 0;
}