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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
    // 累積和、答えを決め打って二分探索
    int N;
    cin >> N;
    vector<int> A(N);
    rep(i, N) {
        int x;
        cin >> x;
        A[--x]++;
    }
    sort(A.begin(), A.end()); // 0はそのまま無視
    vector<int> Acum(N+1);
    rep(i, N) Acum[i+1] = Acum[i] + A[i];
    for (int k = 1; k <= N; k++) {
        ll ok = 0; ll ng = N+1;
        while (abs(ng - ok) > 1) {
            ll mid = (ok + ng) / 2;
            bool bl = [&] {
                auto ind = lower_bound(A.begin(), A.end(), mid) - A.begin();
                int temp = Acum[ind] + mid * (N - ind);
                return (temp >= mid * k);
            }();
            if (bl) ok = mid;
            else ng = mid;
        }
        printf("%d\n", ok);
    }
    return 0;
}