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;

struct Point {
    double x, y;

    Point(double x = 0, double y = 0) : x(x), y(y) {}
};

typedef Point Vector;

Vector operator-(Point A, Point B) {
    return Vector(A.x + B.x, A.y + B.y);
}

Vector operator+(Vector A, Vector B) {
    return Vector(A.x + B.x, A.y + B.y);
}

Vector operator*(Vector A, double p) {
    return Vector(A.x * p, A.y * p);
}

Vector operator/(Vector A, double p) {
    return Vector(A.x / p, A.y / p);
}




const double eps = 1e-7;

int dcmp(double x) {
    if (fabs(x) < eps) return 0;
    else
        return x < 0 ? -1 : 1;
}

bool operator<(const Point &a, const Point &b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

bool operator==(const Point &a, const Point &b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

struct Circle {
    Point c;
    double r;

    Circle() {};

    Circle(Point c, double r) : c(c), r(r) {}

    Point point(double a) {
        return Point(c.x + cos(a) * r, c.y + sin(a) * r);
    }
} cir[100];

Point a[100];
double c[100];

int n, k;

double dis(Point a, Point b) {
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

vector<pair<double, double> > cal(double x1, double y1, double r1, double x2, double y2, double r2) {//两圆交点
    x1 -= x2, y1 -= y2;
    double S = x1 * x1 + y1 * y1, a = (S + r2 * r2 - r1 * r1) / 2, D = S * r2 * r2 - a * a;
    if (D < 0) return {};
    double A1 = a * x1, B1 = y1 * sqrt(D);
    double A2 = a * y1, B2 = x1 * sqrt(D);
    return {{(A1 + B1) / S + x2, (A2 - B2) / S + y2},
            {(A1 - B1) / S + x2, (A2 + B2) / S + y2}};
}

bool check(double t) {
    for (int i = 1; i <= n; i++) {
        cir[i].c = a[i];
        cir[i].r = t / c[i];
    }
    vector<Point> p;
    for (int i = 1; i <= n; i++) {
        p.emplace_back(a[i]);
        for (int j = 1; j < i; j++) {
            // getCircleCircleIntersection(cir[i], cir[j], p);
            auto it = cal(cir[i].c.x, cir[i].c.y, cir[i].r, cir[j].c.x, cir[j].c.y, cir[j].r);
            for (auto v:it)
                p.emplace_back(Point(v.first, v.second));
        }
    }

    for (Point v:p) {
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            if (cir[i].r + eps >= dis(v, cir[i].c)) cnt++;
        }
        if (cnt >= k) return true;
    }
    return false;
}

int main() {
    scanf("%d %d", &n, &k);
    for (int i = 1; i <= n; i++) {
        scanf("%lf %lf %lf", &a[i].x, &a[i].y, &c[i]);
    }

    double l = 0, r = 1e9;
    while (r - l > eps) {
        double mid = (r + l) / 2;
        if (check(mid)) r = mid;
        else l = mid;
    }
    printf("%.6f\n", l);
    return 0;
}