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 <algorithm>
#include <iostream>
using namespace std;

int main()
{
    int n;
    cin >> n;
    int a[300004];
    int d[300004] {0};
    for(int i = 0; i < n; i++){
        cin >> a[i];
        d[a[i]]++;
    }
    sort(d, d + 300004, greater<int>());
    int m = 0;
    for(; m < n; m++) if(d[m] == 0) break;
//    cout << m << endl;
    int rsum[300004];
    rsum[0] = d[0];
    for(int i = 1; i < n; i++) rsum[i] = rsum[i - 1] + d[i];
    for(int k = 1; k <= n; k++){
        if(k > m){
            cout << 0 << endl;
            continue;
        }
        int now = 524288;
        int mid = 262144;
        while(mid >= 1){
            if(k * now - rsum[k - 1] > n - rsum[k - 1]) now -= mid;
            else now += mid;
            mid /= 2;
        }
        if(k * now - rsum[k - 1] > n - rsum[k - 1]) now--;
        cout << now << endl;
    }
}