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.

Examples

Input

3
2 1 2


Output

3
1
0


Input

5
1 2 3 4 5


Output

5
2
1
1
1


Input

4
1 3 3 3


Output

4
1
0
0
#include<bits/stdc++.h>
using namespace std;
//map<int,int> mp;
const int maxn = 3e5 + 10;
int a[maxn], ans[maxn], b[maxn];
priority_queue<int> Q;
int n, m;
void solve() {
	scanf("%d", &n);
	for (int i = 1, x; i <= n; ++i) {
		scanf("%d", &x);
		++a[x];
	}
	for (int i = 1; i <= n; ++i) {
		++b[a[i]];
	}
	for (int i = n; i > 1; --i) {
		b[i - 1] = b[i - 1] + b[i];
	}
	for (int i = 1, sum; i <= n; ++i) {
		sum = 0;
		for (int j = 1; j <= n / i; ++j) {
			sum += b[j];
			if (sum >= j * i)ans[i] = j;
		}
	}
	for (int i = 1; i <= n; ++i)
		printf("%d\n", ans[i]);
}
int main() {
	solve();

	return 0;
}