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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

typedef long long ll;

int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for(int i = 0; i < n; i++) {
    int A;
    cin >> A;
    A--;
    a[A]++;
  }

  sort(a.begin(),a.end());
  vector<int> sum;
  sum = a;

 

  for(int i = 1; i < a.size(); i++) {
    sum[i] += sum[i-1];
  }

 
  for(ll i = 1; i <= n; i++) {
    ll l = 0, r = n + 1;
   
    while(r - l > 1) {
      ll mid = (l + r) / 2;
      ll tmp = upper_bound(a.begin(),a.end(),mid) - a.begin();
      ll x = sum[tmp-1];
      ll y = (a.size() - tmp) * mid;
      //  cout <<"mid = "<<  mid <<"   tmp = "<< tmp <<"   x+y = "<<x+y<< endl;
      if(x + y >= mid * i) l = mid;
      else r = mid;
    }

     cout << l << endl;
  }
}






      
