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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define reps(i, n) for (int i = 1; i <= (n); ++i)
#define repd(i, n) for (int i = (n - 1); i >= 0; --i)
#define repds(i, n) for (int i = (n); i > 0; --i)
#define loop(i, x, n) for (int i = (x); i < (n); ++i)
#define loops(i, x, n) for (int i = (x); i <= (n); ++i)
#define loopd(i, x, n) for (int i = (x); i > (n); --i)
#define loopds(i, x, n) for (int i = (x); i >= (n); --i)
#define itrep(i, a) for (auto i = begin(a); i != end(a); ++i)
#define itrepd(i, a) for (auto i = rbegin(a); i != rend(a); ++i)
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using ld = long double;
template <class T>
using priority_queuel = priority_queue<T, vector<T>, greater<T>>;
constexpr ll inf = static_cast<ll>(1e18);
constexpr int iinf = static_cast<int>(1e9);
constexpr double dinf = 1e10;
constexpr ld ldinf = 1e18;
constexpr double eps = 1e-9;
ll mod = static_cast<ll>(1e9 + 7);
template <class T>
constexpr int sz(const T &a) { return (int)a.size(); }
template <class T>
constexpr T pow2(const T a) { return a * a; }
template <class T, class... Args>
constexpr void mins(T &a, Args... args) { a = min<T>({a, args...}); }
template <class T, class... Args>
constexpr void maxs(T &a, Args... args) { a = max<T>({a, args...}); }
template <class A, int N, class T>
void fillArray(A (&array)[N], const T &val) { std::fill((T *)array, (T *)(array + N), val); }

int n, m;
ll a[300005];
ll s[300005];
bool check(ll k, ll t) {
    ll sum = 0;
    auto idx = lower_bound(a, a + m, t) - a;
    sum += s[idx] + t * (m - idx);
    return sum >= k * t;
}

ll cnt[300005];
void solve() {
    cin >> n;
    rep(i, n) {
        int x;
        cin >> x;
        ++cnt[x - 1];
    }

    rep(i, n) if (cnt[i] > 0) a[m++] = cnt[i];
    sort(a, a + m);
    rep(i, m) s[i + 1] += s[i] + a[i];

    reps(k, n) {
        ll l = 0, r = inf;
        while (r - l > 1) {
            auto mid = (l + r) / 2;
            if (check((ll)k, mid))
                l = mid;
            else
                r = mid;
        }
        cout << l << endl;
    }
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    solve();
    return 0;
}