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 <vector>
#include <complex>
#include <cstdio>

using namespace std;

using P = complex<double>;
using Circle = pair<P,double>;

const double EPS = 1e-12;

bool overlap(const Circle& c1, const Circle& c2){
	auto d = abs(c1.first - c2.first);
	return c2.second - (d + c1.second) > -EPS;
}

bool inCircle(const Circle& c, const P& p){
	return c.second - abs(c.first-p) > -EPS;
}

bool crossCircle(const Circle& c1, const Circle& c2){
	auto r1 = c1.second;
	auto r2 = c2.second;
	auto d = abs(c1.first-c2.first);
	if(d < r1 || d < r2) return d - abs(r1 - r2) > EPS;
	return r1 + r2 - d > -EPS;
}

vector<P> crossPoint(const Circle& c1, const Circle& c2){
    if(!crossCircle(c1, c2)) return vector<P>();
	auto r1 = c1.second;
	auto r2 = c2.second;
	auto r3 = abs(c1.first - c2.first);
	auto rc = (r3*r3 + r1*r1 - r2*r2) / (2*r3);
	auto rs = sqrt(r1*r1 - rc*rc);
	P dif = (c2.first-c1.first)/r3;
    vector<P> res;
    res.emplace_back(c1.first + dif * P(rc,rs));
    res.emplace_back(c1.first + dif * P(rc,-rs));
	return res;
}

int main(){
    int N, K;
    while(cin >> N >> K){
        vector<P> vp;
        vector<int> c(N);
        for(int i=0;i<N;i++){
            int x, y;
            cin >> x >> y >> c[i];
            vp.emplace_back(x, y);
        }
        double L = 0.0, R = 1000000.0;
        for(int _=0;_<200;_++){
            double mid = 0.5*(L+R);
            vector<Circle> vc;
            for(int i=0;i<N;i++) vc.emplace_back(vp[i], mid/c[i]);
            int mcnt = 0;
            for(int i=0;i<N;i++){
                vector<P> check;
                check.push_back(vp[i]);
                for(int j=i+1;j<N;j++){
                    auto cp = crossPoint(vc[i], vc[j]);
                    copy(cp.begin(), cp.end(), back_inserter(check));
                }
                for(auto& cr : check){
                    int cnt = 0;
                    for(int k=0;k<N;k++){
                        if(inCircle(vc[k], cr)) ++cnt;
                    }
                    mcnt = max(mcnt, cnt);
                }
            }
            if(mcnt >= K) R = mid;
            else L = mid;
        }
        printf("%.8lf\n", 0.5*(L+R));
    }
}