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: Accepted)

sum[3<<17],cnt[3<<17];
N,i,acc,t,k,x;
char f[10];
char buf[2200000];
sz;
main()
{
	read(0,buf,2200000);
	for(;;)
	{
		k=buf[sz++];
		if(k<'0')break;
		N=N*10+k-'0';
	}
	for(i=0;i<N;i++)
	{
		x=0;
		for(;;)
		{
			k=buf[sz++];
			if(k<'0')break;
			x=x*10+k-'0';
		}
		++sum[cnt[x]++];
	}
	i=sz=0;
	for(t=N;t;t--)
	{
		while(acc>=i*t)acc+=sum[i++];
		cnt[t]=i-1;
	}
	for(i=1;i<=N;i++)
	{
		if(!cnt[i])
		{
			buf[sz++]='0';
		}
		else
		{
			if(cnt[i-1]!=cnt[i])
			{
				k=0;
				t=cnt[i];
				while(t)
				{
					f[k++]=t%10;
					t/=10;
				}
				acc=k;
			}
			else k=acc;
			while(k--)buf[sz++]=f[k]+'0';
		}
		buf[sz++]='\n';
		if(sz>100000)
		{
			write(1,buf,sz);
			sz=0;
		}
	}
	write(1,buf,sz);
}