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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define int long long
#define inf 1000000007
#define LINF 100000000000000007LL
#define ll long long
using namespace std;
int ans[314514],sum[314514];
signed main(){
	int n;
	cin>>n;
	vector<int> a,u;
	map<int,int> mope;
	for(int i=0;i<n;i++){
		int x;
		cin>>x;
		a.pb(x);
		u.pb(x);
		mope[x]++;
	}
	sort(u.begin(),u.end());
	u.erase( unique( u.begin(),u.end() ), u.end() );
	vector<int> cnt;
	cnt.pb(0);
	for(int i=0;i<u.size();i++){
		cnt.pb( mope[ u[i] ] );
	}
	sort(cnt.begin(),cnt.end());
	for(int i=1;i<=cnt.size();i++){
		sum[i] = sum[i-1] + cnt[i];
	}
	int cur = 0;
	for(int i=n;i>=1;i--){
		if( i >= cnt.size() ) ans[i] = 0;
		else{
			int start = cnt.size() - i;
			int ng = cnt.size(), ok = start; // start ~ ok can be same
			while( ng - ok > 1 ){
				int mid = ( ok + ng ) / 2;
				int num = mid - start + 1;
				int tmp = cnt[mid] * num - ( sum[mid] - sum[start-1] );
				if( tmp > cur ){
					ng = mid;
				}
				else{
					ok = mid;
				}
			}
			int amari = cur - cnt[ok] * (ok-start+1) + ( sum[ok] - sum[start-1] );
			ans[i] = cnt[ok] + amari/(ok-start+1);
			cur += cnt[start];
		}
	}
	for(int i=1;i<=n;i++){
		cout<<ans[i]<<endl;
	}
	return 0;
}