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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

uses Math;
var
	N,K,i,j:Longint;
	x,y,c:Array[1..60]of Longint;
	rr:Array[1..60]of double;
	L,R,M,a,t,d,ri,rj,yy:double;
	flag:Boolean;
function count(tx,ty:double):Boolean;
var i,cnt:Longint;
begin
	cnt:=0;
	for i:=1 to N do if sqr(X[i]-tx)+sqr(Y[i]-ty)<sqr(rr[i])+0.000000001 then begin
		inc(cnt);
		if cnt=K then begin
			count:=True;
			exit;
		end;
	end;
	count:=False;
end;
begin
	read(N,K);
	for i:=1 to N do read(x[i],y[i],c[i]);
	L:=0;
	R:=200000;
	while R-L>0.000001 do begin
		M:=(L+R)/2;
		for i:=1 to N do rr[i]:=M/c[i];
		flag:=False;
		for i:=1 to N do begin
			if count(X[i],Y[i])then begin
				flag:=True;
				break;
			end;
		end;
		if not flag then begin
			for i:=2 to N do begin
				for j:=1 to i-1 do begin
					d:=sqr(X[i]-X[j])+sqr(Y[i]-Y[j]);
					ri:=rr[i];
					rj:=rr[j];
					yy:=4*sqr(ri)*d-sqr(sqr(ri)+d-sqr(rj));
					if yy<0 then continue;
					a:=arctan2(sqrt(yy),sqr(ri)+d-sqr(rj));
					t:=arctan2(Y[j]-Y[i],X[j]-X[i]);
					if count(X[i]+ri*cos(t+a),Y[i]+ri*sin(t+a)) then flag:=True
					else if count(X[i]+ri*cos(t-a),Y[i]+ri*sin(t-a))then flag:=True;
					if flag then break;
				end;
				if flag then break;
			end;
		end;
		if flag then R:=M else L:=M;
	end;
	writeln(R:0:9);
end.