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 Ruby (2.3.3) (judge verdict: Accepted)

def max(a,b) a>b ? a:b end

def distance(x1, y1, x2, y2)
  dx, dy = x1 - x2, y1 - y2
  Math.sqrt(dx**2 + dy**2)
end

# 時間t, 熱源位置(x,y)で焼ける肉の個数を返す
# t + 0.0000000001 の 0.0000000001 はtan,cosの計算による誤差をカバー(だと思う)
def yaketa_niku(t, x, y)
  res = 0
  N.times do |i|
    if t + 0.0000000001 >= C[i] * distance(x, y, X[i], Y[i])
      res += 1
    end
  end
  res
end

# [二円の交点] http://godfoot.world.coocan.jp/circle-circle.htm
def circle_cross_point(t, xyc1, xyc2)
  x1, y1, x2, y2 = xyc1[0], xyc1[1], xyc2[0], xyc2[1]
  d = distance(x1, y1, x2, y2)
  r1, r2 = t/xyc1[2], t/xyc2[2]

  return if d > r1 + r2 || d < (r1 - r2).abs
  
  tmp = (r1**2 + d**2 - r2**2)/(2*r1*d)
  return unless -1 <= tmp && tmp <= 1
  
  sita = Math.atan2(y2-y1, x2-x1)
  alf = Math.acos(tmp)

  nx1 = x1 + r1 * Math.cos(sita + alf)
  ny1 = y1 + r1 * Math.sin(sita + alf)
  nx2 = x1 + r1 * Math.cos(sita - alf)
  ny2 = y1 + r1 * Math.sin(sita - alf)
  [[nx1,ny1], [nx2,ny2]].uniq
end

# ci*(熱源からの距離) <= t
# (熱源からの距離) <= t/ci
# 時間tの時、肉が焼ける円(中心は肉の座標、半径はt/ci)があり、
# 2つの円を選んだ時の交点の座標を熱源の候補位置を計算する
# 計算した熱源の位置を使って肉が時間内に焼けた個数の最大値を計算し返す
def f(t)
  heater_pos = XY
  test=[]
  XYC_comb.each do |xyc1, xyc2|
    tmp = circle_cross_point(t, xyc1, xyc2)
    heater_pos += tmp if tmp
  end
  res = 0
  heater_pos.each do |x, y|
    res = max(res, yaketa_niku(t, x, y))
  end
  res
end

#時間で二分探索
def main()
  l = 0.0
  r = 400000.0
  100.times do
    m = (l+r)/2.0
    if f(m) < K
      l = m
    else
      r = m
    end
    break if r-l < 10**(-7)
  end
  puts sprintf("%.10f", l)
end

N, K = gets.split.map(&:to_i)
XYC = N.times.map{gets.split.map(&:to_f)}

=begin
N=60
K=30
aaa=Array.new(N)
N.times do |i|
  aaa[i] = [-1000 + 30*i, -1000 + 30*i, rand(1..100)].map(&:to_f)
end
XYC=aaa
=end

X, Y, C = XYC.transpose
XY = [X, Y].transpose
XYC_comb = XYC.combination(2).to_a

main