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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::{self, BufRead};
use std::collections::{BinaryHeap, BTreeMap};
use std::cmp::Ordering;

#[derive(PartialEq, Eq, Debug, Copy, Clone, Default, Hash)]
struct Reverse(usize);

impl PartialOrd for Reverse {
    fn partial_cmp(&self, other: &Reverse) -> Option<Ordering> {
        other.0.partial_cmp(&self.0)
    }
    fn lt(&self, other: &Self) -> bool {
        other.0 < self.0
    }
    fn le(&self, other: &Self) -> bool {
        other.0 <= self.0
    }
    fn gt(&self, other: &Self) -> bool {
        other.0 > self.0
    }
    fn ge(&self, other: &Self) -> bool {
        other.0 >= self.0
    }
}

impl Ord for Reverse {
    fn cmp(&self, other: &Reverse) -> Ordering {
        other.0.cmp(&self.0)
    }
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let n: usize = lines.next().unwrap().unwrap().parse().unwrap();
  let xs: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  let mut m: BTreeMap<usize, usize> = BTreeMap::new();
  for x in xs {
    let v = 1 + m.get(&x).unwrap_or(&0);
    m.insert(x, v);
  }
  let mut xs: BinaryHeap<Reverse> = BinaryHeap::new();
  for (_, v) in m {
    xs.push(Reverse(v));
  }
  let mut ys: Vec<usize> = vec![0; n];
  let l = xs.len();
  for i in 0..l {
    let j = l - i - 1;
    if let Some(Reverse(x)) = xs.pop() {
      ys[j] = x;
      if let Some(Reverse(y)) = xs.pop() {
        xs.push(Reverse(x + y));
      }
    }
  }
  for y in ys {
    println!("{}", y);
  }
}