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 int li;
typedef vector<li> vi;
typedef pair<li,li> pi;
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define B begin()
#define E end()
#define f(i,l,r) for(li i=l;i<=r;++i)
#define fr(i,l,r) for(li i=l;i>=r;--i)
#define u_map unordered_map

li fastpow(li base,li exp,li M) 
{
    li res=1;
    while(exp>0) {
        if(exp&1)
            res=(res*base)%M;

        base=(base*base)%M;
        exp>>=1;
    }
    return res;
}
    
int main()
{    
    ios_base::sync_with_stdio(false);
        cin.tie(NULL);

        li n;

        cin>>n;

        li a[n];
        u_map<li,li> m;

        for(li i=0;i<n;i++)
        {
            cin>>a[i];
            m[a[i]]++;
        }

        priority_queue<li,vi,greater<li>> pq;

        for(auto it:m)
        {
            pq.push(it.S);
        }

        vi ans(n+1);

        f(i,1,n)
        {
            while(pq.size()>=i)
            {
                ans[i]+=pq.top();

                li temp=pq.top();
                pq.pop();
                li tt=i;

                stack<li> s;

                while(--tt)
                {
                    s.push(pq.top());pq.pop();
                }

                while(!s.empty())
                {
                    if(s.top()-temp>0)
                    pq.push(s.top()-temp);

                    s.pop();
                }
            }

            while(!pq.empty())
                pq.pop();

            for(auto it:m)
            {
                pq.push(it.S);
            }

            cout<<ans[i]<<endl;
        }

    return 0;
}