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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	static int count(double xs[] , double ys[] , double C[] , double sec , double x , double y){
		int cnt = 0;
		for(int i = 0 ; i < xs.length ; ++i){
			double d = Math.hypot(xs[i] - x, ys[i] - y);
			if(C[i] * d <= sec){
				++cnt;
			}			
		}
		return cnt;
	}
	// sec秒後に焼ける肉の枚数がK個以上かどうか判定する
	static boolean can(double xs[] , double ys[] , double C[] , double sec, int K){
		for(int i = 0 ; i < xs.length ; ++i){
			if(count(xs , ys , C , sec, xs[i] , ys[i]) >= K){
				return true;
			}
			for(int j = 0 ; j < xs.length ; ++j){
				double D = Math.hypot(xs[i] - xs[j], ys[i] - ys[j]);
				double ri = sec / C[i];
				double rj = sec / C[j];
				if(D > ri + rj){
					continue;
				}
				if(ri > D + rj || rj > D + ri){
					continue;
				}
				double a = Math.acos((ri * ri + D * D - rj * rj) / (2 * ri * D));
				double t = Math.atan2(ys[j] - ys[i], xs[j] - xs[i]);
				if(count(xs, ys, C, sec, xs[i] + Math.cos(t + a) * ri, ys[i] + Math.sin(t + a) * ri) >= K){
					return true;
				}
				if(count(xs, ys, C, sec, xs[i] + Math.cos(t - a) * ri, ys[i] + Math.sin(t - a) * ri) >= K){
					return true;
				}
			}
		}
		return false;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		double xs[] = new double[N];
		double ys[] = new double[N];
		double C[] = new double[N];
		for(int i = 0 ; i < N ; ++i){
			xs[i] = sc.nextDouble();
			ys[i] = sc.nextDouble();
			C[i] = sc.nextDouble();
		}
		double low = 0.0;
		double high = 0.0;
		for(int i = 0 ; i < N ; ++i){
			double d = Math.hypot(xs[i], ys[i]) * C[i];
			high = Math.max(high, d);		
		}
		while((high - low) >= 1.0e-12){
			double mid = (high + low) / 2;
			if(can(xs, ys, C, mid, K)){
				high = mid;
			}else{
				low = mid;
			}
		}
		System.out.println(low);
	}
}