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;

const double EPS = 1e-8;

struct Point{
    double x, y, c;
};

vector<pair<double, double>> intersection(double x1, double y1, double r1, double x2, double y2, double r2){
    double ab = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));

    if (-EPS < (r1 - r2) - ab || ab - (r1 + r2) > -EPS)
        return {};
    
    double ad = (r1 * r1 - r2 * r2 + ab * ab) / (2 * ab);
    double x0 = x1 + (x2 - x1) * ad / ab;
    double y0 = y1 + (y2 - y1) * ad / ab;
    double n_x = -(y2 - y1), n_y = x2 - x1;
    double norm = sqrt((n_x * n_x) + (n_y * n_y));
    double cd = sqrt((r1 * r1) - (ad * ad));

    return {{x0 + n_x * cd / norm, y0 + n_y * cd / norm}, {x0 - n_x * cd / norm, y0 - n_y * cd / norm}};
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    cout << fixed << setprecision(10);

    int n, k;
    cin >> n >> k;

    vector<Point> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i].x >> a[i].y >> a[i].c;

    auto C = [&](double mid){
        vector<double> r(n);
        for (int i = 0; i < n; i++)
            r[i] = mid / a[i].c;

        vector<pair<double, double>> center;
        for (int i = 0; i < n; i++){
            center.push_back({a[i].x, a[i].y});
            for (int j = i + 1; j < n; j++){
                auto v = intersection(a[i].x, a[i].y, r[i], a[j].x, a[j].y, r[j]);

                for (auto k: v)
                    center.push_back(k);
            }
        }
        
        for (auto c: center){
            int cnt = 0;
            for (auto i: a){
                if (mid - i.c * sqrt((c.first - i.x) * (c.first - i.x) + (c.second - i.y) * (c.second - i.y))  > -EPS)
                    cnt += 1;
            }

            if (cnt >= k)
                return true;
        }

        return false;

    };

    double lb = 0.0, ub = 1e9;
    for (int i = 0; i < 100; i++){
        double mid = (lb + ub) / 2;
        if (C(mid))
            ub = mid;
        else
            lb = mid;
    }

    cout << ub << '\n';
    
    return 0;
}