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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <assert.h>
typedef int64_t ll;
typedef uint64_t ull;
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)

#define MAXN (300001)
#define MOD (1000000007)

// 指定された要素以上の値が現れる最初の位置のイテレータ
// [l,r)
typedef int lb;
int lower_bound(lb* a, lb val, int l, int r)
{
    if(l==r) return l;
    int m = (l+r)/2;
    if(a[m]<val) return lower_bound(a,val,m+1,r);
    else return lower_bound(a,val,l,m);
}

int c[MAXN];
int cs[MAXN];
int main(void)
{
    ll n;
    scanf("%ld",&n);
    for(int i=0;i<n;i++)
    {
        ll a;
        scanf("%ld",&a);
        c[a]++;
    }
    qsort(c,MAXN,sizeof(int),acs);
    cs[0] = c[0];
    for(int i=1;i<MAXN;i++) cs[i]=cs[i-1]+c[i];
    
    for(int k=1;k<=n;k++)
    {
        int ans = n/k;
        for(;ans>0;ans--)
        {
            int i = lower_bound(c,ans,0,MAXN);
            int s = i>0 ? cs[i-1] : 0;
            s += (MAXN-i)*ans;
            if(s>=(ans*k)) break;
        }
        printf("%d\n", ans);
    }
    return 0;
}