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;
using Int = long long;
int PREP = (cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(9), 0);
//int SEGV = getenv("D") || (exit(system("D= SEGFAULT_SIGNALS=all catchsegv ./prog.exe") >> 8), 0);
using Pt = complex<double>;
vector<Pt> circleCircle(Pt o1, double r1, Pt o2, double r2) {
    Pt d = o2 - o1;
    double d2 = norm(d);
    if (d2 == 0) return {};
    double pd = (d2 + r1*r1 - r2*r2) / 2;
    double h2 = r1*r1 - pd*pd/d2;
    if (h2 >= 0) {
        Pt p = o1 + d*pd/d2;
        Pt h = d * Pt(0, 1) * sqrt(h2/d2);
        return {p-h, p+h};
    }
    return {};
}
int main() {
    int N, K; cin >> N >> K;
    vector<int> X(N), Y(N), C(N);
    for (int i = 0; i < N; i++) cin >> X[i] >> Y[i] >> C[i];
    auto f = [&](double t) {
        for (int i = 0; i < N; i++) {
            int cnt = 0;
            Pt p = Pt(X[i], Y[i]);
            for (int k = 0; k < N; k++) {
                Pt c = Pt(X[k], Y[k]);
                if (abs(c - p) <= t / C[k]) cnt++;
            }
            if (cnt >= K) return true;
        }
        for (int i = 0; i < N; i++) {
            for (int j = i + 1; j < N; j++) {
                Pt a = Pt(X[i], Y[i]);
                Pt b = Pt(X[j], Y[j]);
                vector<Pt> pts = circleCircle(a, t / C[i], b, t / C[j]);
                for (auto p : pts) {
                    int cnt = 0;
                    for (int k = 0; k < N; k++) {
                        if (k == i || k == j) cnt++;
                        else {
                            Pt c = Pt(X[k], Y[k]);
                            if (abs(c - p) <= t / C[k]) cnt++;
                        }
                    }
                    if (cnt >= K) return true;
                }
            }
        }
        return false;
    };
    double ok = 1e9, ng = 0;
    for (int i = 0; i < 100; i++) {
        double mid = (ok + ng) / 2;
        (f(mid) ? ok : ng) = mid;
    }
    cout << ok << endl;
    return 0;
}