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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n, i, j, temp, br=0;
    scanf("%d", &n);
    int a[n+1];
    for(i=0; i<n; i++){
		scanf("%d", &a[i]);
    }

    for(i=0; i<n; i++){
		for(j=i+1; j<n; j++){
			if(a[i]>a[j]){
				temp=a[i];
				a[i]=a[j];
				a[j]=temp;
			}
		}
    }

    for(i=0; i<n; i++){
		if(a[i]!=a[i+1]){
			br++;
		}
    }

    printf("%d\n", n);
    for(i=2; i<=n; i++){
		if(br/i>0){
			printf("%d\n", br/i);
		}
		else printf("0\n");
    }


    return 0;
}