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)

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() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let mut c = vec![0; n + 1];
    for _ in 0..n {
        let a: usize = scan(tokens);
        c[a] += 1;
    }
    let mut d = vec![0; n + 1];
    for i in 0..n + 1 {
        d[c[i]] += 1;
    }
    let mut cumsum0 = vec![0; n + 1];
    let mut cumsum1 = vec![0; n + 1];
    for i in 1..n + 1 {
        cumsum0[i] = cumsum0[i - 1] + d[i];
        cumsum1[i] = cumsum1[i - 1] + d[i] * i;
    }
    let mut i = n;
    for k in 1..n + 1 {
        while i > 0 && (cumsum1[i] + (cumsum0[n] - cumsum0[i]) * i) / i < k {
            i -= 1;
        }
        println!("{}", i);
    }
}