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 <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

int main(){
  int N;
  cin >> N;
  vector<int> B(N);
  for(int i = 0; i < N; ++i){
    int a;
    cin >> a;
    --a;
    ++B[a];
  }
  sort(B.begin(), B.end());
  for(auto b : B) cerr << b << " ";
  cerr << endl;
  priority_queue<int> pq;
  for(int i = 0; i < N; ++i)
    pq.push(-B[i]);
  vector<int> ans(N);
  for(int i = N-1; i >= 0; --i){
    ans[i] = -pq.top();
    // cerr << i << " " << -pq.top() << endl;
    if(!i) break;
    int t = -pq.top();
    pq.pop();
    while(t){
      int k = pq.top();
      pq.pop();
      --k, --t;
      pq.push(k);
    }
  }
  for(int i = 0; i < N; ++i)
    cout << ans[i] << endl;
}
