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: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int nums = sc.nextInt();
        int[] list = new int[nums];
        for(int i = 0; i < nums; i++){
            int num =  sc.nextInt();
            list[i]= num;
        }
        Arrays.sort(list);
        System.out.println(nums);
        for (int k= 2; k <= nums ; k++) {
            int[] copied = Arrays.copyOfRange(list, 0, list.length);
            int count = eatCards(k, copied);
            Arrays.sort(list);
            System.out.println(count);
        }
    }

    public static int eatCards(int k, int[] list){
        int counter = 0;
        for (int j = 0; j < list.length; j++) {
            int num = k;
            if(list[j]!=-1){
                int first = list[j];
                list[j] = -1;
                num--;
                for(int i= j + 1; i < list.length; i++) {
                    if (first != list[i]) {
                        num--;
                        first = list[i];
                        list[i] = -1;
                        if(num == 0){
                            counter++;
                            break;
                        }
                    }
                }
            }
        }
        return  counter;
    }
}