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: Accepted)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
//#include<bits/stdc++.h>
#include<cstdio>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

//いやーコンテスト中は絶対無理
int hist[300002], cnt[300001];
const int cm = 1 << 17;
char cn[cm], * ci = cn + cm, ct;
inline char getcha() {
	if (ci - cn == cm) { fread_unlocked(cn, 1, cm, stdin); ci = cn; }
	return *ci++;
}
inline int getint() {
	int A = 0;
	if (ci - cn + 16 > cm) while ((ct = getcha()) >= '0') A = A * 10 + ct - '0';
	else while ((ct = *ci++) >= '0') A = A * 10 + ct - '0';
	return A;
}
const int dm = 1 << 21;
char dn[dm], * di = dn;
inline void putint(int X) {
	int keta = 0;
	char C[10];
	while (X) {
		*(C + keta) = '0' + X % 10;
		X /= 10;
		keta++;
	}
	for (int i = keta - 1; i >= 0; i--)* di++ = (*(C + i));
	*di++ = '\n';
}

int main() {
	//cin.tie(0);
	//ios::sync_with_stdio(false);


	int N = getint();
	rep(i, N) hist[getint()]++;
	rep1(i, N) cnt[hist[i]]++;

	int ruiseki = N;
	int mae = 1;
	int kei = 1;
	int ijou = 0;
	for (int i = N; i >= 1; i--) {
		ijou += cnt[i];
		ruiseki -= cnt[i] * i;
		while ((kei - ijou) * i <= ruiseki) kei++;

		for (int j = mae; j < kei; j++) putint(i);
		mae = kei;
	}
	for (int j = mae; j <= N; j++) {
		*di++ = '0';
		*di++ = '\n';
	}
	fwrite(dn, 1, di - dn, stdout);

	Would you please return 0;
}