Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 3
-1 0 3
0 0 3
1 0 2
1 1 40

Sample Output 1

2.4

If we put the heat source at \left(-0.2, 0\right), the 1-st, 2-nd, and 3-rd pieces of meat will be ready to eat within 2.4 seconds. This is the optimal place to put the heat source.

Sample Input 2

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

Sample Output 2

7411.2252

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// F
// DeP
#pragma GCC optimize(3, "inline", "Ofast")
#include <ctime>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;

namespace IO {
    const int MAXSIZE = 1 << 18 | 1;
    char buf[MAXSIZE], *p1, *p2;

    inline int Gc() { return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin), p1 == p2)? EOF: *p1++; }
    template<typename T> inline void read(T& x) {
        x = 0; int f = 0, ch = Gc();
        while (!isdigit(ch)) f |= ch == '-', ch = Gc();
        while (isdigit(ch)) x = x * 10 + ch - '0', ch = Gc();
        if (f) x = -x;
    }
}
using IO::read;

const int MAXN = 65;
const double TIME_LIMIT = 1.99, DELTA = 0.91, GG = 1.995;
const double EPS = 1e-9;

namespace Geo {
    struct Vector {
        double x, y;
        Vector(double _x = 0.0, double _y = 0.0): x(_x), y(_y) { }
    };
    typedef Vector Point;
}
using namespace Geo;

int n, K;
int C[MAXN];
Point P[MAXN], Ans;
double A[MAXN], ans;

inline double metaphysics() { return (double(rand()) / RAND_MAX) * 2.0 - 1.0; }

inline double update(const Point& p) {
    for (int i = 1; i <= n; ++i)
        A[i] = C[i] * sqrt((P[i].x - p.x) * (P[i].x - p.x) + (P[i].y - p.y) * (P[i].y - p.y));
    sort(A+1, A+1+n);
    const double& now = A[K];
    if (now < ans) Ans = p, ans = now;
    return now;
}

inline void SA() {
    double T = 3422.0;
    double nx = Ans.x, ny = Ans.y, now = ans;
    while (T > EPS) {
		double nxtx = nx + T * metaphysics(), nxty = ny + T * metaphysics();
        if (clock() > GG * CLOCKS_PER_SEC) printf("%.6lf\n", ans), exit(0);
        double delta = update(Point(nxtx, nxty)) - now;
		if (exp(-delta / T) > rand()) nx = nxtx, ny = nxty;
        T *= DELTA;
    }
}

int main() {
    srand( time(nullptr) );
    read(n), read(K);
    for (int i = 1; i <= n; ++i) {
        static int x, y;
        read(x), read(y), read(C[i]);
        P[i] = Point(x, y), Ans.x += x, Ans.y += y;
    }
    Ans.x /= n, Ans.y /= n;
    ans = update(Ans);
    while (clock() < TIME_LIMIT * CLOCKS_PER_SEC) SA();
    printf("%.6lf\n", ans);
    return 0;
}