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 Ruby (2.3.3) (judge verdict: Runtime Error)

N def max(a,b); a > b ? a : b; end
  
N = gets.to_i
A = gets.split.map(&:to_i)
C = A.each_with_object(Array.new(N + 1, 0)){|i,c| c[i] += 1 }
D = C.each_with_object(Array.new(N + 1, 0)){|j,d| d[j] += 1 }
E = Array.new(N + 1, 0)
(1 .. N).each{|k| E[k] = E[k - 1] + k * D[k] }
F = Array.new(N + 1, 0)
(1 .. N).each{|k| F[k] = F[k - 1] + D[k] }

def max_size(m)
  (E[m] + m * (F[N] - F[m])) / m
end

dp = Array.new(N + 1, 0)
(1 .. N).each{|m| dp[m] = max_size(m) }
G = Array.new(N + 1, 0)

dp.each_with_index do |k,m|
  G[k] = m
end

(1 .. N).reverse_each.inject(0) do |s,i|
  G[i] = max(s, G[i])
end

puts G[1 .. N]