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)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn main() {
    use std::collections::HashMap;
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let mut map = HashMap::new();
    for _ in 0..n {
        let a: usize = scan(tokens);
        *map.entry(a).or_insert(0) += 1;
    }
    let t: Vec<usize> = {
        let mut count = Vec::with_capacity(map.len());
        for (_, c) in map {
            count.push(c);
        }
        count.sort();
        count.reverse();
        let mut t = Vec::with_capacity(count[0] + 1);
        t.push(0);
        let mut j = 0;
        for i in (1..count[0] + 1).rev() {
            while j < count.len() && count[j] >= i {
                j += 1;
            }
            t.push(j);
        }
        t
    };
    println!("{}", n);
    for k in 2..n + 1 {
        let mut i = 0;
        let mut j = t.len();
        while i + 1 < j {
            let c = (i + j) / 2;
            if t[c] < k {
                i = c;
            } else {
                j = c;
            }
        }
        let mut ans = 0;
        let mut tmp = 0;
        while j < t.len() {
            ans += (t[j] + tmp) / k;
            tmp = (t[j] + tmp) % k;
            j += 1;
            while t[i] + tmp >= k {
                ans += 1;
                tmp = (t[i] + tmp) % k;
                i -= 1;
            }
        }
        println!("{}", ans);
    }
}