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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>

using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)


using ld = long double;
using Point =  complex<ld>;
const ld eps = 1e-7;

//円
class Circle {
public:
	Point p;
	ld r;
};

vector<Point> is_cc(Circle c1,Circle c2){
    if(c1.r>c2.r)swap(c1,c2);
    ld d = abs(c1.p-c2.p);
    if(d>c1.r+c2.r){
        return {};
    }
    if(c2.r>d+c1.r)return {};
    ld theta = acosl((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));
    Point mid = c1.p +(c2.p-c1.p)* cosl(theta) * c1.r /d;
    Point h = (c2.p-c1.p) * Point(0,1) * sinl(theta) * c1.r /d;
    vector<Point> ret;
    ret.push_back(mid+h);ret.push_back(mid-h);
    return ret;
}

void solve(){
    int n,K;
    cin>>n>>K;
    vector<Point> p(n);
    vector<ld> a(n);
    rep(i,n){
        int x,y;cin>>x>>y;
        p[i]=Point(x,y);
        cin>>a[i];
    }

    ld ok = 1e6, ng = 0;
    rep(_,100){
        ld mid = (ok+ng)/2;
        vector<Circle> c(n);
        rep(i,n)c[i]={p[i],mid/a[i]};
        int cnt  = 0;
        rep(i,n)rep(j,i){
            auto is = is_cc(c[i],c[j]);
            for(auto e:is){
                int ret = 0;
                rep(k,n)if(abs(e-c[k].p)<=c[k].r || k==i || k==j)++ret;
                cnt = max(cnt,ret);
            }
        }
        rep(i,n){
            int ret = 0;
            rep(j,n)if(abs(c[i].p-c[j].p)<=c[j].r)++ret;
            cnt = max(cnt,ret);
        }
        if(cnt>=K)ok=mid;
        else ng=mid;
    }
    cout<<ok<<endl;
}
int main(){
    cout<<fixed<<setprecision(12);
    solve();
    return 0;
}