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)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

use std::cmp::{max, min};
use std::collections::BTreeMap;

fn main() {
    input! {
        n: usize,
        a: [i64; n],
    }

    let mut mm = BTreeMap::new();
    for a in a {
        *mm.entry(a).or_insert(0) += 1;
    }

    let mut v: Vec<usize> = mm.values().cloned().collect();
    v.sort();
    v.reverse();

    let mut psum: Vec<usize> = vec![0; v.len() + 1];
    for i in 0..v.len() {
        psum[i + 1] = psum[i] + v[i];
    }

    let mut acc = 0;

    let mut anss = vec![];

    for i in (0..v.len()).rev() {
        let mut l = 0;
        let mut r = i;

        {
            let chk = |m: usize| -> bool {
                let rect = v[m] * (i - m + 1);
                let sub = psum[i + 1] - psum[m];
                rect <= sub + acc
            };

            while r - l > 1 {
                let m = (l + r) / 2;
                if chk(m) {
                    r = m;
                } else {
                    l = m;
                }
            }

            if chk(l) {
                r = l;
            }

            let l = r;

            let w = i - l + 1;
            let area = psum[i + 1] - psum[l] + acc;
            let h = (area - (w - 1)) / w;
            let ans = if l > 0 { min(v[l - 1], h) } else { h };
            let ans = max(ans, v[i]);
            anss.push(ans);
        }
        acc += v[i];
    }

    anss.reverse();

    for ans in anss.iter() {
        println!("{}", ans);
    }

    for _ in anss.len()..n {
        println!("0");
    }
}