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 LuaJIT (2.0.4) (judge verdict: Wrong Answer)

local mma, mmi = math.max, math.min
local msq = math.sqrt
local n, k = io.read("*n", "*n")
local t = {}
for i = 1, n do
  t[i] = {io.read("*n", "*n", "*n")}
end

local function func(x, y)
  local q = {}
  for i = 1, n do
    q[i] = t[i][3] * t[i][3] * ((t[i][1] - x) * (t[i][1] - x) + (t[i][2] - y) * (t[i][2] - y))
  end
  table.sort(q)
  return msq(q[k])
end

local function solve(tbl)
  local xmin, xmax, ymin, ymax = tbl[1], tbl[2], tbl[3], tbl[4]
  local dv = 96
  local dxrange = (xmax - xmin) / (dv * 2)
  local dyrange = (ymax - ymin) / (dv * 2)
  local mpx, mpy, minval = false, false, 0
  for i = 1, dv do
    local xp = xmin + dxrange * (i * 2 - 1)
    yp = ymin - dyrange
    for j = 1, dv do
      yp = yp + dyrange * 2
      local v = func(xp, yp)
      if not mpx or v < minval then
        minval = v
        mpx, mpy = i, j
      end
    end
  end
  tbl[1] = xmin + mma(0, dxrange * (mpx * 2 - 1 - dv / 2))
  tbl[2] = mmi(xmax, xmin + dxrange * (mpx * 2 - 1 + dv / 2))
  tbl[3] = ymin + mma(0, dyrange * (mpy * 2 - 1 - dv / 2))
  tbl[4] = mmi(ymax, ymin + dyrange * (mpy * 2 - 1 + dv / 2))
end

local tbl = {-1000, 1000, -1000, 1000}
local rep = 10000000
while true do
  solve(tbl)
  if 1.84 < os.clock() then break end
end
local xm, ym = (tbl[1] + tbl[2]) / 2, (tbl[3] + tbl[4]) / 2
local ret = func(xm, ym)
print(string.format("%.10f", ret))