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.*;

class Main{
	public static void main(String[] $){
		System.err.println();
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		int[]d=new int[n];
		for(int i=0;i<n;++i)
			++d[s.nextInt()-1];

		Arrays.sort(d);
		//System.err.println(Arrays.toString(d));

		/*
		int[]count=new int[n+1];
		for(int i:d)
			++count[i];
		System.err.println(Arrays.toString(count));
		*/

		int[]p=new int[n+1];
		System.arraycopy(d,0,p,1,n);
		Arrays.parallelPrefix(p,Integer::sum);
		//System.err.println(Arrays.toString(p));

		for(int k=1;k<=n;++k)
			System.out.println((p[n+1-k]+k-1)/k);
	}
}