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: Time Limit Exceeded)

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

fn calc(m: usize, cnt: &[usize]) -> usize {
    let mut h = std::collections::BinaryHeap::new();
    for c in cnt {
        h.push(*c);
    }
    let mut s = Vec::with_capacity(m);
    let mut ans = 0;
    while h.len() >= m {
        ans += 1;
        for _ in 0..m {
            let v = h.pop().unwrap();
            if v > 1 {
                s.push(v - 1);
            }
        }
        for &v in &s {
            h.push(v);
        }
        s.clear();
    }
    ans
}

fn eval(l: usize, r: usize, x: usize, y: usize, cnt: &[usize], ans: &mut [usize]) {
    if l >= r {
        return;
    }
    if x + 1 == y {
        for i in l..r {
            ans[i] = x;
        }
        return;
    }
    let m = (l + r) / 2;
    ans[m] = calc(m, cnt);
    eval(l, m, ans[m], y, cnt, ans);
    eval(m + 1, r, x, ans[m] + 1, cnt, ans);
}

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.sort();
    cnt.reverse();
    while let Some(c) = cnt.pop() {
        if c > 0 {
            cnt.push(c);
            break;
        }
    }
    let mut ans = vec![0; n + 1];
    eval(1, cnt.len() + 1, 0, n + 1, &cnt, &mut ans);
    for a in &ans[1..] {
        writeln!(out, "{}", *a).ok();
    }
}

fn main() {
    run();
}