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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

package com.abc143;

import java.util.Arrays;
import java.util.Scanner;
import java.util.function.Function;

public class Main {
    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int n = std.nextInt();
        Integer[] a = new Integer[n];
        for (int i = 0; i < n; i++) {
            a[i] = 0;
        }
        for (int i = 0; i < n; i++) {
            int x = std.nextInt() - 1;
            a[x]++;
        }

        Integer[] sorted = Arrays.stream(a).sorted().toArray(Integer[]::new);
        int[] s = new int[n + 1];
        for (int i = 0; i < n; i++) {
            s[i + 1] = s[i] + sorted[i];
        }

        // 1 ~ n takes
        for (int k = 1; k <= n; k++) {
            final int kIndex = k;
            int l = 0, r = n/k + 1;
            while (l + 1 < r) {
                int c = (l + r) / 2;
                Function<Integer, Boolean> f = integer -> {
                    int i = ~Arrays.binarySearch(sorted, integer, (l1, r1) -> l1.compareTo(r1) >= 0 ? 1 : -1);
                    int sum = integer * (n - i) + s[i];
                    return (sum >= integer * kIndex);
                };

                if (f.apply(c)) {
                    l = c;
                } else {
                    r = c;
                }
            }

            System.out.println(l);
        }

    }
}