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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

void merge_sort(int x[], int n)
{
	static int y[300001] = {};
	if (n <= 1) return;
	merge_sort(&(x[0]), n/2);
	merge_sort(&(x[n/2]), (n+1)/2);
	
	int i, p, q;
	for (i = 0, p = 0, q = n/2; i < n; i++) {
		if (p >= n/2) y[i] = x[q++];
		else if (q >= n) y[i] = x[p++];
		else y[i] = (x[p] < x[q])? x[p++]: x[q++];
	}
	for (i = 0; i < n; i++) x[i] = y[i];
}

int main ()
{
	int i, N, A, count[300001];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) {
		scanf("%d", &A);
		count[A]++;
	}
	merge_sort(&(count[1]), N);
	
	int j, k, l, sum[300001] = {}, ans, tmp;
	for (i = 1, sum[0] = 0; i <= N; i++) sum[i] = sum[i-1] + count[i];
	for (i = 1, l = N; i <= N; i++) {
		ans = sum[l] / (l - N + i);
		if (sum[N-i+1] * (l - N + i) <= sum[l]) {
			ans = sum[N-i+1];
			l = N - i + 1;
		}
		printf("%d\n", ans);
	}
	fflush(stdout);
	return 0;
}