human
Can you provide a Python program to determine whether Dmitry can make the MEX of the array equal to exactly i given the following:
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. 
Solve the task in PYTHON3.
assistant
#Here is Python program to determine whether Dmitry can make the MEX of the array equal to exactly i.

import array
import bisect
import heapq
import math
import collections
import sys
import copy
from functools import reduce
import decimal
from io import BytesIO, IOBase
import os
import itertools
import functools
from types import GeneratorType

#
# sys.setrecursionlimit(10 ** 9)
decimal.getcontext().rounding = decimal.ROUND_HALF_UP

graphDict = collections.defaultdict

queue = collections.deque

################## pypy deep recursion handling ##############
# Author = @pajenegod
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        to = f(*args, **kwargs)
        if stack:
            return to
        else:
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        return to
                    to = stack[-1].send(to)

    return wrappedfunc

################## Graphs ###################
class Graphs:
    def __init__(self):
        self.graph = graphDict(set)

    def add_edge(self, u, v):
        self.graph[u].add(v)
        self.graph[v].add(u)

    def dfs_utility(self, nodes, visited_nodes, colors, parity, level):
        global count
        if nodes == 1:
            colors[nodes] = -1
        else:
            if len(self.graph[nodes]) == 1 and parity % 2 == 0:
                if q == 1:
                    colors[nodes] = 1
                else:
                    colors[nodes] = -1
                    count += 1
            else:
                if parity % 2 == 0:
                    colors[nodes] = -1
                else:
                    colors[nodes] = 1
        visited_nodes.add(nodes)
        for neighbour in self.graph[nodes]:
            new_level = level + 1
            if neighbour not in visited_nodes:
                self.dfs_utility(neighbour, visited_nodes, colors, level - 1, new_level)

    def dfs(self, node):
        Visited = set()
        color =...