description: Dmitry has an array of n non-negative integers a_1, a_2, ..., a_n.
  
  In one operation, Dmitry can choose any index j (1 ≤ j ≤ n) and increase the value of the element a_j by 1. He can choose the same index j multiple times.
  
  For each i from 0 to n, determine whether Dmitry can make the MEX of the array equal to exactly i. If it is possible, then determine the minimum number of operations to do it.
  
  The MEX of the array is equal to the minimum non-negative integer that is not in the array. For example, the MEX of the array [3, 1, 0] is equal to 2, and the array [3, 3, 1, 4] is equal to 0.
  
  Input
  
  The first line of input data contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. 
  
  The descriptions of the test cases follow.
  
  The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the array a.
  
  The second line of the description of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ n) — elements of the array a.
  
  It is guaranteed that the sum of the values n over all test cases in the test does not exceed 2⋅10^5.
  
  Output
  
  For each test case, output n + 1 integer — i-th number is equal to the minimum number of operations for which you can make the array MEX equal to i (0 ≤ i ≤ n), or -1 if this cannot be done.
  
  Example
  
  Input
  
  
  5
  3
  0 1 3
  7
  0 1 2 3 4 3 2
  4
  3 0 0 0
  7
  4 6 2 3 5 0 5
  5
  4 0 1 0 4
  
  
  Output
  
  
  1 1 0 -1 
  1 1 2 2 1 0 2 6 
  3 0 1 4 3 
  1 0 -1 -1 -1 -1 -1 -1 
  2 1 0 2 -1 -1 
  
  Note
  
  In the first set of example inputs, n=3:
  
    * to get MEX=0, it is enough to perform one increment: a_1++; 
    * to get MEX=1, it is enough to perform one increment: a_2++; 
    * MEX=2 for a given array, so there is no need to perform increments; 
    * it is impossible to get MEX=3 by performing increments. 
solution: import os, sys
  import unittest.mock
  from io import BytesIO, IOBase
  import math
  import sys
  import collections
  from bisect import bisect
  from operator import xor
  from functools import reduce
  from bisect import bisect
  import itertools
  import sys
  from collections import Counter
  from math import ceil, log2
  from functools import reduce
  from itertools import starmap, chain
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  class Mod:
      def __init__(self, mod=10e9 + 7):
          self.mod = mod
  
      def add(self, a, b):
          return (a + b) % self.mod
  
      def mul(self, a, b):
          return (a * b) % self.mod
  
      def div(self, a, b):
          return self.mul(a, self.inv(b))
  
      def inv(self, a):
          if a == 1:
              return 1
          return ((self.mod - self.mod // a) * self.inv(self.mod % a)) % self.mod
  
      def binpow(self, a, b):
          ans = 1
          while b > 0:
              if b % 2 == 1:
                  ans = self.mul(ans, a)
              a = self.mul(a, a)
              b //= 2
          return ans
  
  
  class Dsu:
      def __init__(self, n):
          self.p = [i for i in range(n)]
          self.count = n
          self.size = [1] * n
  
      def root(self, x):
          if x == self.p[x]:
              return x
          else:
              self.p[x] = self.root(self.p[x])
              return self.p[x]
  
      def merge(self, x, y):
          r1 = self.root(x)
          r2 = self.root(y)
          if r1 == r2:
              return r1
          if self.size[r2] > self.size[r1]:
              r1, r2 = r2, r1
          self.size[r1] += self.size[r2]
          self.p[r2] = r1
          self.count -= 1
          return r1
  
      def sizee(self, r):
          return self.size[self.root(r)]
  
      def count(self):
          return self.count
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  L = lambda: sys.stdin.readline().rstrip("\r\n")
  N = lambda: int(L())
  F = lambda: float(L())
  Ns = lambda: list(map(int, L().split()))
  Ss = lambda: list(L().split())
  Nss = lambda: sorted(Ns())
  Fs = lambda: list(map(float, L().split()))
  
  
  def gcd(*args):
      return reduce(math.gcd, args, 0)
  
  
  def presolve():
      pass
  
  def mygcd(a, b):
      """Calculate the Greatest Common Divisor of a and b.
  
      Unless b==0, the result will have the same sign as b (so that when
      b is divided by it, the result comes out positive).
      """
      while b:
          a, b = b, a%b
      return a
  # solution is here
  # sys.setrecursionlimit(10 ** 9)
  multitest = True
  def solve():
      n = N()
      a = Ns()
      cnt = [0] * (n + 1)
      for x in a:
          cnt[x] += 1
      stack = []
      res = []
      add = 0
      spaces = True
      for i in range(n + 1):
          if cnt[i] > 0 and spaces:
              res.append(str(cnt[i] + add))
              for j in range(cnt[i] - 1):
                  stack.append(i)
          else:
              if spaces:
                  res.append(str(cnt[i] + add))
              else:
                  res.append(str(-1))
              if len(stack) > 0:
                  add += i - stack.pop()
                  cnt[i] = 1
              else:
                  spaces = False
      print(" ".join(res))
  
  
  
  
  
  
      pass
  # dp = [[0] * n for _ in range(n)]
  # path = [[0] * n for _ in range(n)]
  # for i in range(n):
  #     dp[i][i] = a[i]
  # # dp[i][j] = min(dp[i][k] + dp[k + 1][j])
  # for ln in range(1, n):
  #     for i in range(n - ln):
  #         left = i
  #         right = i + ln
  #         dp[left][right] = 2 ** 65
  #         for k in range(i, right):
  #             left_sum = dp[left][k]
  #             right_sum = dp[k + 1][right]
  #             if left_sum + right_sum <= dp[left][right]:
  #                 dp[left][right] = left_sum + right_sum
  #                 path[left][right] = k
  # print(dp[0][n - 1])
  
  def exceptSolve():
      try:
          solve()
      except Exception as e:
          print(e)
          raise
  
  
  if multitest:
      presolve()
      t = N()
      while t != 0:
          exceptSolve()
          t -= 1
  else:
      exceptSolve()
  