human
Give me a Python code 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. 
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python code to determine whether Dmitry can make the MEX of the array equal to exactly i.

import os
import sys
from io import BytesIO, IOBase
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')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

from collections import Counter

def solve(n, a):
	c = Counter(a)
	ans = [-1 for i in range(n + 1)]
	base = [None for i in range(n + 1)]
	curops = [None for i in range(n + 1)]

	curops[0] = c[0]
	ans[0] = c[0]
	
	excess_lessvals = []
	## base setup
	if c[0] == 0:
		return ' '.join(map(str,ans))
	else:
		base[0] = 0
		if c[0] > 1:
			excess_lessvals.append([0,...