Problem:
Score : 600 points

Problem 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.

Constraints

1 \le N \le 3 \times 10^5

1 \le A_i \le N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print N integers.
The t-th (1 \le t \le N) of them should be the answer for the case K=t.

Sample Input 1

3
2 1 2

Sample Output 1

3
1
0

For K = 1, we can do the operation as follows:

Choose the first card to eat.

Choose the second card to eat.

Choose the third card to eat.

For K = 2, we can do the operation as follows:

Choose the first and second cards to eat.

For K = 3, we cannot do the operation at all. Note that we cannot choose the first and third cards at the same time.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

5
2
1
1
1

Sample Input 3

4
1 3 3 3

Sample Output 3

4
1
0
0

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let scan_int () = Scanf.scanf " %d" (fun x -> x)
let scan_array len read_func = Array.init len (fun _ -> read_func())
let print_int n = Printf.printf "%d\n" n

let input n =
  let arr = Array.make n 0 in
  let _ =
    for i = 0 to n-1 do
      let a = scan_int() in
      arr.(a-1) <- arr.(a-1) + 1
    done in
  arr

let cumulative_sum n arr =
  let ret = Array.make (n+1) 0 in
  let _ =
    for i = 0 to n-1 do
      ret.(i+1) <- ret.(i) + arr.(i)
    done in
  ret

let rec binary_search left right eval =
  if right - left <= 1 then right
  else 
    let mid = (right + left) / 2 in
    if eval mid then binary_search mid right eval
    else binary_search left mid eval

let inner_eval arr value mid = arr.(mid) > value

let eval i arr cum remain mid =
  let r = binary_search (-1) i (inner_eval arr mid) in
  let require = (i-r) * mid - (cum.(i) - cum.(r)) in
  remain >= require

let solve i n arr cum =
  let remain = n - cum.(i) in
  (binary_search 0 (n+1) (eval i arr cum remain)) - 1  (* get left *)


let () =
  let n = scan_int() in
  let arr = input n in
  let _ = Array.fast_sort (fun x y -> if x < y then 1 else 0) arr in
  let cum = cumulative_sum n arr in
  for i = 1 to n do
    print_int (solve i n arr cum)
  done