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: Time Limit Exceeded)

#pragma region RegionDefs
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include "boost/multi_array.hpp"
#define rep(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
#define reps(i,j,n) for(int i=(j),i##_len=(n);i<i##_len;++i)
#define repr(i,n) for(int i=(n);i>=0;--i)
#define all(x) (x).begin(),(x).end()
using namespace std;
template<class T=int> using P  = pair<T, T>;
template<class T=int> using V  = vector<T>;
template<class T=int> using PQ = priority_queue<T>;
template<class T=int> using PQG = priority_queue<T, V<T>, greater<T>>;
typedef long long ll;
const int INF = 1e9;
const ll MOD = 1000000007LL;
void in() {}
template<class Head, class... Tail>
void in(Head&& head, Tail&&... tail)
{
    cin >> head;
    in(move(tail)...);
}
#define IN(...) int __VA_ARGS__; in(__VA_ARGS__)
#define VIN(T, v, n) V<T> v(n); for(auto& _elem:v)cin>>_elem
struct UnionFind
{
    V<int> par;
    V<int> siz;
    UnionFind(int n) : par(n), siz(n, 1)
    {
        rep(i, n) par[i] = i;
    }
    int root(int x)
    {
        while (par[x] != x) x = par[x] = par[par[x]];
        return x;
    }
    bool unite(int x, int y)
    {
        int rx = root(x);
        int ry = root(y);
        if (rx == ry) false;
        if (siz[rx] < siz[ry]) swap(rx, ry);
        siz[rx] += siz[ry];
        par[ry] = rx;
        return true;
    }
    int same(int x, int y)
    {
        return root(x) == root(y);
    }
};

#pragma endregion RegionDefs

void solve()
{
    IN(n);
    VIN(int, v, n);
    sort(all(v));
    int a = v[0];
    int count = 0;
    PQ<int> q;
    rep(i, n)
    {
        if (v[i] == a) count++;
        else
        {
            q.push(count);
            count = 1;
            a = v[i];
        }
    }
    if (count > 0) q.push(count);
    reps(i, 1, n+1)
    {
        int ans = 0;
        auto q_ = q;
        auto b = new int[i];
        while (true)
        {
            rep(j, i)
            {
                if (q_.empty() || q_.top() <= 0) goto end;
                b[j] = q_.top()-1;
                q_.pop();
            }
            rep(j, i)
            {
                q_.push(b[j]);
            }
            ans++;
        }
        end:;
        cout << ans << endl;
    }
}

int main()
{
    cin.tie(0); cout.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(numeric_limits<double>::max_digits10);

    solve();

    return 0;
}