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
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define pii pair<int,int>
#define pli pair<ll,int>
#define pil pair<int,ll>
#define pll pair<ll,ll>
#define cpd complex<double>
const cpd rot90(0, 1);
const int inf = 1 << 30;
const ll linf = 1e18;
const double EPS = 1e-7;
template<class T> void chmin(T& x, T y){if(x > y) x = y;}
template<class T> void chmax(T& x, T y){if(x < y) x = y;}

struct circle{
    complex<double> c;//中心
    double r;//半径
    circle(complex<double> c, double r) : c(c), r(r) {}
};

vector<complex<double>> culc_intersection(circle c1, circle c2){
    double l = abs(c1.c - c2.c), r1 = c1.r, r2 = c2.r;
    vector<complex<double>> ps;

    if(l > r1 + r2) return ps;
    if(r1 > l + r2) return ps;
    if(r2 > l + r1) return ps;

    double x = ((r1 * r1) - (r2 * r2) + (l * l)) / (2 * l);
    double a = sqrt((r1 * r1) - (x * x));
    complex<double> vec1 = (c2.c - c1.c) * x / l;
    complex<double> vec2 = (c2.c - c1.c) * rot90 * a / l;

    ps.push_back(c1.c + vec1 + vec2);
    if(vec2 != complex<double>(0, 0)) ps.push_back(c1.c + vec1 - vec2);
    return ps;
}

bool culc_eps(db x, db y){
    return abs(x - y) > EPS * (abs(x) + abs(y));
}

struct meat{
    cpd p;
    db c;
    meat(cpd p, db c) : p(p), c(c) {}
};

int N, K;
vector<meat> f; 
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> N >> K;
    for(int i = 0; i < N; i++){
        double x, y, c;
        cin >> x >> y >> c;
        f.push_back(meat{cpd(x, y), c});
    }
    double lb = 0, ub = 1e6;
    while(culc_eps(lb, ub)){
        db T = (ub + lb) / 2;
        vector<circle> cs;
        for(int i = 0; i < N; i++){
            cs.push_back(circle{f[i].p, T / f[i].c});
        }

        vector<cpd> ps;
        for(int i = 0; i < N; i++){
            for(int j = i; j < N; j++){
                if(i == j){
                    ps.push_back(f[i].p);
                }else{
                    vector<cpd> qs = culc_intersection(cs[i], cs[j]);
                    for(auto& u : qs) ps.push_back(u);
                }
            }
        }

        bool flag = false;
        for(auto u : ps){
            int cnt = 0;
            for(int i = 0; i < N; i++){
                double l = abs(u - cs[i].c);
                if(l <= cs[i].r *(1 + EPS)) cnt++;
            }
            if(cnt >= K) flag = true;
        }
        if(flag) ub = T;
        else lb = T;
    }
    cout << fixed << setprecision(15);
    cout << ub << endl;
    return 0;
}