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;
int freq[1000006], res[1000006];
priority_queue<int>pq;
queue<int>q;
void func(int x)
{
    int k= 0;
    while(k<x)
    {
        q.push(pq.top()-1);
        pq.pop();
        k++;
    }
    while(!q.empty())
    {
        if(q.front())pq.push(q.front());
        q.pop();
    }
    return;
}
int main()
{
    int n, x;
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    scanf("%d", &x), freq[x]++;
    for(int i=1; i<=n; i++)
    if(freq[i])pq.push(freq[i]);
    x= pq.size();
    int tot= 0;
    while(pq.size()>=x)
    {
        func(x);
        tot+= x;
    }
    res[x]= (tot/x);
    for(int i=x-1; i>0; i--)
    {
        res[i]= (tot/i);
        int rem= tot%i;
        if(rem)
        {
            rem= i-rem;
            if(pq.size()>=rem)
            func(rem), tot+= rem, res[i]++;
        }
        while(pq.size()>=i)func(i), res[i]++, tot+= i;
    }
    for(int i=1; i<=n; i++)
    printf("%d\n", res[i]);
    return 0;
}