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: Accepted)

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;

struct Point {
	long double px, py;
};

struct Circle {
	long double px, py, r;
};

long double PI = 3.14159265358979;
long long N, K;
Point G[1 << 18]; long double C[1 << 18];
Circle H[1 << 18];

double dst(Point V1, Point V2) {
	return sqrt((V1.px - V2.px) * (V1.px - V2.px) + (V1.py - V2.py) * (V1.py - V2.py));
}

double dst(Circle V1, Circle V2) {
	return sqrt((V1.px - V2.px) * (V1.px - V2.px) + (V1.py - V2.py) * (V1.py - V2.py));
}

int solve(double T) {
	for (int i = 1; i <= N; i++) {
		H[i].px = G[i].px;
		H[i].py = G[i].py;
		H[i].r = T / C[i];
	}

	vector<Point> Major;
	for (int i = 1; i <= N; i++) {
		Major.push_back(Point{ H[i].px, H[i].py });
	}
	for (int i = 1; i <= N; i++) {
		for (int j = i + 1; j <= N; j++) {
			if (dst(H[i], H[j]) > H[i].r + H[j].r) continue;
			
			double DST = dst(H[i], H[j]);
			double DX = (H[j].px - H[i].px) / DST;
			double KAKUDO = acos(DX);
			if (H[j].py < H[i].py) KAKUDO = PI * 2.0 - KAKUDO;

			double cl = 0.0, cr = PI, cm;
			for (int k = 0; k < 50; k++) {
				cm = (cl + cr) / 2.0;
				double vx = H[i].px + H[i].r * cos(KAKUDO + cm);
				double vy = H[i].py + H[i].r * sin(KAKUDO + cm);
				double ds = dst(Circle{ vx, vy, 0.0 }, H[j]);
				if (ds < H[j].r) { cl = cm; }
				else { cr = cm; }
			}

			double tx = H[i].px + H[i].r * cos(KAKUDO + cm);
			double ty = H[i].py + H[i].r * sin(KAKUDO + cm);
			double ux = H[i].px + H[i].r * cos(KAKUDO - cm);
			double uy = H[i].py + H[i].r * sin(KAKUDO - cm);
			Major.push_back(Point{ tx, ty });
			Major.push_back(Point{ ux, uy });
		}
	}

	int maxn = 0;
	for (int i = 0; i < Major.size(); i++) {
		int cnt = 0;
		for (int j = 1; j <= N; j++) {
			double I = dst(G[j], Major[i]);
			if (I <= H[j].r + 1e-7) cnt++;
		}
		maxn = max(maxn, cnt);
	}
	return maxn;
}

int main() {
	cin >> N >> K;
	for (int i = 1; i <= N; i++) cin >> G[i].px >> G[i].py >> C[i];

	long double L = 0, R = 1e6, M;
	for (int i = 0; i < 50; i++) {
		M = (L + R) / 2.0;
		int I = solve(M);
		if (I >= K) { R = M; }
		else { L = M; }
	}
	printf("%.12Lf\n", M);
	return 0;
}