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: Wrong Answer)

n = gets.chomp.to_i
sum = 0
idxsmap = {}
a = []
gets.chomp.split(' ').map{|ii|
  sum += 1
  i = ii.to_i
  if idxsmap[i].nil?
    a.push(1)
    idx = a.length - 1
    idxsmap[i] = idx
  else
    a[idxsmap[i]] += 1
  end
}
a.sort!{|i, j| j <=> i }

# p a

sums = [sum]

(n - 1).times{|ii|
  k = ii + 2
  af = a.clone

  count = 0
  while af.length >= k
    (af.length / k).times{|jj|
      j = jj + 1
      willeat = af[k * j - 1]
      k.times{|i|
        af[k * (j - 1) + i] -= willeat
      }
      count += willeat

      # p '-- eat --'
      # p af
    }

    # p '--- done ---'
    # p af
    af -= [0]
    af.sort!{|i, j| j <=> i }
    # p af
  end

  sums[k - 1] = count
}

puts sums