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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#define rep(i,n) for(ll i=0;i<n;++i)
#define rep1(i,n) for(ll i=1;i<=n;++i)
using namespace std;
typedef long long ll;
ll n;
ll m;
vector<ll> a;
vector<ll> sum;
vector<ll> rle;
vector<ll> v;
bool C(ll x,ll y){
  return y<v[x];
}
ll upper(ll l,ll r,ll y){//条件を満たす中で最小のものを返す
  ll lb=l-1,ub=r+1;
  while(ub-lb>1){
    ll mid=(lb+ub)/2;
    if(C(mid,y)) ub=mid;
    else lb=mid;
  }
  return ub;
}
bool C2(ll y,ll k){
  ll j=upper(0,m-1,y);
  return sum[m]-sum[j]-(m-j)*y<=n-y*k;
}

ll lower(ll l,ll r,ll k){
  
  ll lb=l-1,ub=r+1;
  while(ub-lb>1){
    ll mid=(lb+ub)/2;
    if(C2(mid,k)) lb=mid;
    else ub=mid;
  }
  return lb;
}

int main()
{
  cin >> n;
  a.resize(n);
  rep(i,n) cin >> a[i];

  sort(a.begin(), a.end());


  ll cnt=1;
  rep(i,n-1){
    if(a[i]==a[i+1]){
      cnt++;
    }
    else{
      rle.push_back(cnt);
      cnt=1;
      v.push_back(a[i]);
    }
  }
  rle.push_back(cnt);
  v.push_back(a[n-1]);
  
  m=rle.size();
  

  
  sum.resize(m+1);
  sum[0]=0;
  rep(i,m){
    sum[i+1] = sum[i]+rle[i];
  }


  rep1(k,n){
    cout << lower(1,n/k,k) << "\n";
  }
  return 0;
}