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: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int asc(const void *a, const void *b) {
  return *(int *)a - *(int *)b;
}
int desc(const void *a, const void *b) {
  return *(int *)b - *(int *)a;
}

int main(void) {
  int N;
  int i,j,k;
  int *A,*B,*C,*D;
  int var;
  
  scanf("%d\n", &N);
  A = (int *)calloc(N,sizeof(int));
  B = (int *)calloc(N,sizeof(int));
  for(i=0;i<N;i++)
  {
  	scanf("%d", &A[i]);
  }
  //printf("%d %d %d\n",A[0],A[1],A[2]);
  qsort(A, N, sizeof(int), asc);
  //printf("%d %d %d\n",A[0],A[1],A[2]);
  var=1;
  for(i=0;i<N-1;i++)
  {
    B[i]=var;
  if(A[i]!=A[i+1])
   {
   var+=1;
   }
  }
  B[N-1]=var;
  //count
  C = (int *)calloc(var,sizeof(int));
  D = (int *)calloc(var,sizeof(int));
  for(i=0;i<N;i++)
  {
  C[B[i]-1]+=1;
  }
  //printf("B %d %d %d\n",C[0],C[1],C[2]);
  qsort(C, var, sizeof(int), desc);
  //printf("A %d %d %d\n",C[0],C[1],C[2]);
  
  //Cに枚数の配列できてる。
  //ここあら繰り返し。
  int K;
  int dist_num;
  int count;
  
  for(K=1;K<N+1;K++)
  {
  memcpy(D, C, var*sizeof(int));
  dist_num =var;
  count =0;
  while(dist_num >K-1)
  {	
    count ++;
    //printf("%d %d %d\n",D[0],D[1],D[2]);
    qsort(D, var, sizeof(int), desc);
    for(i=0;i<K;i++)
  	{
  		D[i]--;	
	}
    dist_num =0;
    for(i=0;i<var;i++)
    {
    	dist_num +=(D[i]>0?1:0);
    }
  }
  printf("%d\n",count);
  }//K
}