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.

Examples

Input

3
2 1 2


Output

3
1
0


Input

5
1 2 3 4 5


Output

5
2
1
1
1


Input

4
1 3 3 3


Output

4
1
0
0
#include<bits/stdc++.h>
#define maxn 300005
#define LL long long
using namespace std;
int n,m,a[maxn],b[maxn];
LL s[maxn];
bool check(LL x,int k){
	int p=lower_bound(b+1,b+1+m,x)-b;
	return s[p-1]+(m-p+1)*x>=x*k;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	sort(a+1,a+1+n);
	for(int i=1;i<=n;i++)
		if(a[i]!=a[i-1]) b[++m]=1;
		else b[m]++;
	sort(b+1,b+1+m);
	for(int i=1;i<=m;i++) s[i]=s[i-1]+b[i];
	for(int k=1;k<=n;k++){
		LL l=0,r=s[m]/k,mid;
		while(l<r){
			mid=(l+r+1)>>1;
			if(check(mid,k)) l=mid;
			else r=mid-1;
		}
		printf("%lld\n",l);
	}
}