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
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce


sys.setrecursionlimit(2147483647)
INF = 10 ** 20
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007



def bi_search(k):
    ok = 0
    ng = n // k + 1
    while ng > ok + 1:
        mid = (ng + ok) // 2
        if calc(mid, k):
            ok = mid
        else:
            ng = mid
    return ok


def calc(x, k):
    # x回できるか?
    l = bisect_left(L, (x, 0))
    # 全部使えるのが左側、l回しか使えないのが右側。
    if kv_acc[l - 1] + acc[l] * x >= k * x:
        return True
    else:
        return False




n = I()
A = LI()
L = [(0, 0)] + sorted([(k, v) for k, v in Counter(Counter(A).values()).items()])
m = len(L)
acc = [0] * (m + 1)
for i in range(m - 1, -1, -1):
    acc[i] = acc[i + 1] + L[i][1]

kv_acc = list(accumulate([k * v for k, v in L]))
for k in range(1, n + 1):
    print(bi_search(k))


