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)

rd()
{
	int x=0,k;
	for(;;)
	{
		k=getchar_unlocked();
		if(k<'0'||k>'9')break;
		x=x*10+k-'0';
	}
	return x;
}
wt(int x)
{
	int s=0;
	char f[10];
	while(x)
	{
		f[s++]=x%10;
		x/=10;
	}
	if(!s)f[s++]=0;
	while(s--)putchar_unlocked(f[s]+'0');
	putchar_unlocked('\n');
}
sum[3<<17],cnt[3<<17],ans[3<<17];
N,i,acc,t;
main()
{
	N=rd();
	for(i=0;i<N;i++)
	{
		++sum[cnt[rd()]++];
	}
	i=0;
	for(t=N;t>0;t--)
	{
		while(acc>=i*t)acc+=sum[i++];
		ans[t]=i-1;
	}
	for(i=1;i<=N;i++)wt(ans[i]);
}