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 <iomanip>
#include <algorithm>
#include <cmath>
#include <vector>

template <class T>
T sq(T x) { return x * x; }

using ldouble = long double;

struct Point {
    ldouble x, y;
    Point(ldouble x, ldouble y) : x(x), y(y) {}

    Point operator+(const Point& p) const {
        return Point(x + p.x, y + p.y);
    }
};

struct Circle {
    Point p;
    ldouble r;
    Circle(ldouble x, ldouble y, ldouble r)
        : p(x, y), r(r) {}
};

constexpr ldouble EPS = 1e-10;

int compare(ldouble lhs, ldouble rhs) {
    if (std::abs(lhs - rhs) < EPS) {
        return 0;
    } else {
        return lhs < rhs ? -1 : 1;
    }
}

ldouble dist(const Point& a, const Point& b) {
    return std::hypot(a.x - b.x, a.y - b.y);
}

std::vector<Point> crosspoint(
    const Circle& a, const Circle& b) {
    std::vector<Point> ret;

    ldouble d = dist(a.p, b.p);
    if (compare(d, a.r + b.r) > 0 ||
        compare(d, std::abs(a.r - b.r)) < 0) return ret;

    ldouble theta = std::acos((sq(d) + sq(a.r) - sq(b.r)) / (a.r * d * 2));
    ldouble phi = std::atan2(b.p.y - a.p.y, b.p.x - a.p.x);

    ret.emplace_back(a.p.x + a.r * std::cos(phi + theta),
                     a.p.y + a.r * std::sin(phi + theta));
    ret.emplace_back(a.p.x + a.r * std::cos(phi - theta),
                     a.p.y + a.r * std::sin(phi - theta));
    return ret;
}

void solve() {
    int n, k;
    std::cin >> n >> k;

    std::vector<Circle> ps;
    std::vector<ldouble> cs;
    for (int i = 0; i < n; ++i) {
        ldouble x, y, c;
        std::cin >> x >> y >> c;
        ps.emplace_back(x, y, 0);
        cs.push_back(c);
    }

    ldouble ok = 1e9, ng = 0;
    for (int q = 0; q < 100; ++q) {
        ldouble t = (ok + ng) / 2;

        std::vector<Point> qs;
        for (int i = 0; i < n; ++i) {
            ps[i].r = t / cs[i];
            qs.push_back(ps[i].p);
        }

        for (int j = 0; j < n; ++j) {
            for (int i = 0; i < j; ++i) {
                for (auto p : crosspoint(ps[i], ps[j])) {
                    qs.push_back(p);
                }
            }
        }

        bool judge = false;
        for (auto p : qs) {
            int cnt = 0;
            for (auto c : ps) {
                auto d = dist(p, c.p);
                if (compare(d, c.r) <= 0) ++cnt;
            }
            if (cnt >= k) judge = true;
        }

        if (judge) {
            ok = t;
        } else {
            ng = t;
        }
    }

    std::cout << std::fixed << std::setprecision(10)
              << ok << std::endl;
}

int main() {
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    std::ios::sync_with_stdio(false);

    solve();

    return 0;
}