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 JAVA):

Solution:
import java.util.*;

class Main {
	Scanner sc;
	int N, K;
	int[] x, y, c;
	static final double eps = 1e-7;
	
	void calc() {
		sc = new Scanner(System.in);
		N = sc.nextInt();
		K = sc.nextInt();
		
		x = new int[N];
		y = new int[N];
		c = new int[N];
		for (int i = 0; i < N; i++) {
			x[i] = sc.nextInt();
			y[i] = sc.nextInt();
			c[i] = sc.nextInt();
		}
		
		double r = 2000d * Math.sqrt(2) * 100;
		double l = 0d;
		double m = 0;
		while (Math.abs(r-l)/r > eps) {
			m = (l + r) / 2;
			if (!Kintersects(m)) l = m;
			else r = m;
		}
		System.out.println(m);
	}
	
	private boolean Kintersects(double t) {
		for (int i = 0; i < N; i++) {
			if (Kcontains(t, x[i], y[i])) return true;
		}
		for (int i = 0; i < N-1; i++) {
			for (int j = i+1; j < N; j++) {
				double x2, y2, r1, r2, b, a, d, xa, ya;
				x2 = x[j] - x[i];
				y2 = y[j] - y[i];
				r1 = t/c[i];
				r2 = t/c[j];
				b = x2*x2 + y2*y2;
				a = (b + r1*r1 - r2*r2) / 2;
				d = b*r1*r1 - a*a;
				if (d < 0) continue;
				d = Math.sqrt(d);
				xa = (a*x2 + y2*d)/b + x[i];
				ya = (a*y2 - x2*d)/b + y[i];
				if (Kcontains(t, xa, ya)) return true;
				xa = (a*x2 - y2*d)/b + x[i];
				ya = (a*y2 + x2*d)/b + y[i];
				if (Kcontains(t, xa, ya)) return true;
			}
		}
		return false;
	}

	private boolean Kcontains(double t, double x, double y) {
		int c = 0;
		for (int i = 0; i < N; i++) {
			if (contains(i, t, x, y))
				if ( ++c >= K ) return true;
		}
		return false;
	}
	
	private boolean contains(int j, double t, double x, double y) {
		double xs, ys, d;
		xs = this.x[j] - x;
		ys = this.y[j] - y;
		d = Math.sqrt( xs*xs + ys*ys );
		return (d <= t/c[j]*(1+eps));
	}
	
	public static void main(String[] args) {
		new Main().calc();
	}
}