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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using Int = long long;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N; cin >> N;
    map<int, int> M;
    for (int i = 0; i < N; i++) {
        int A; cin >> A; M[A]++;
    }
    vector<int> B;
    for (auto &p : M) B.push_back(p.second);
    sort(begin(B), end(B));
    vector<int> C(N + 1);
    int j = B.size() - 1;
    int sum = N;
    for (int i = N; i > 0; i--) {
        while (j >= 0 && B[j] > i) j--;
        sum -= (B.size() - 1) - j;
        C[i] = sum / i;
    }
    j = N;
    for (int i = 1; i <= N; i++) {
        while (j > 0 && C[j] < i) j--;
        cout << j << '\n';
    }
    return 0;
}