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: Accepted)

use std::io::{Read, Write};

fn run() {
    let out = std::io::stdout();
    let mut out = std::io::BufWriter::new(out.lock());
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let mut cnt = vec![0; n + 1];
    for _ in 0..n {
        let a: usize = it.next().unwrap().parse().unwrap();
        cnt[a] += 1;
    }
    cnt.push(1_000_000_000);
    cnt.sort();
    cnt.reverse();
    let mut sum = cnt.clone();
    sum.push(0);
    for i in (0..cnt.len()).rev() {
        sum[i] += sum[i + 1];
    }
    let mut ans = vec![0; n + 1];
    let mut p = n;
    for i in (1..(n + 1)).rev() {
        p = std::cmp::min(i, p);
        while cnt[p - 1] * (i - p + 1) <= sum[p] {
            p -= 1;
        }
        ans[i] = sum[p] / (i - p + 1);
    }
    for a in &ans[1..] {
        writeln!(out, "{}", *a).ok();
    }
}

fn main() {
    run();
}