output_description: Print N integers. The t-th (1 \le t \le N) of them should be the answer for
  the case K=t.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      A_1 A_2 \ldots A_N
attempt: import bisect
  
  n = int(input())
  k = sorted(map(int, input().split()))
  
  hist = []
  count = 0
  for i in range(n):
      count += 1
      if i == n - 1 or k[i] != k[i + 1]:
          hist.append(count)
          count = 0
  hist.sort()
  
  sumh = [0]
  for i in range(len(hist)):
      sumh.append(sumh[-1] + hist[i])
  
  
  def ok(h, d):
      i = bisect.bisect_left(hist, h)
      s = sumh[i] + h * (len(hist) - i)
      return s >= h * d
  
  
  def binary_search(d):
      lo, hi = 0, n // d + 1
      while hi - lo > 1:
          mid = (lo + hi) // 2
          if ok(mid, d):
              lo = mid
          else:
              hi = mid
      return lo
  
  
  for d in range(1, n + 1):
      print(binary_search(d))
  
problem_description: Statement
  
  Takahashi has N cards. The i-th of these cards has an integer A_i written on
  it.
  
  Takahashi will choose an integer K, and then repeat the following operation
  some number of times:
  
    * Choose exactly K cards such that the integers written on them are all different, and eat those cards. (The eaten cards disappear.)
  
  For each K = 1,2, \ldots, N, find the maximum number of times Takahashi can do
  the operation.
samples: [{"input": "3\n    2 1 2", "output": "3\n    1\n    0\n    \n\nFor K = 1, we can do the operation as follows:\n\n  * Choose the first card to eat.\n  * Choose the second card to eat.\n  * Choose the third card to eat.\n\nFor K = 2, we can do the operation as follows:\n\n  * Choose the first and second cards to eat.\n\nFor K = 3, we cannot do the operation at all. Note that we cannot choose the\nfirst and third cards at the same time.\n\n* * *"}, {"input": "5\n    1 2 3 4 5", "output": "5\n    2\n    1\n    1\n    1\n    \n\n* * *"}, {"input": "4\n    1 3 3 3", "output": "4\n    1\n    0\n    0"}]