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 <bits/stdc++.h>

using namespace std;

typedef long long ll;

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    vector<int> a(n + 1, 0);
    for (int i = 0; i < n; ++i) {
        int x;
        cin >> x;
        a[x]++;
    }
    vector<int> b(n + 1, 0);
    for (int i = 0; i < n; ++i) {
        b[a[i]]++;
    }
    int last = n;
    int s = n;
    for (int k = 1; k <= n; ++k) {
        while (last > s / k) {
            s -= b[last];
            b[last - 1] += b[last];
            last--;
        }
        cout << last << '\n';
    }
    return 0;
}