human
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. 

impo
assistant
#Here is Python code to determine whether Dmitry can make the MEX of the array equal to exactly i.
rt sys, collections, math, bisect, heapq, random, functools,io,os,copy
input = sys.stdin.readline
out = sys.stdout.flush
def rep():
    a = list(map(int,input().split()))
    return a

def sep():
    a = list(input().split())
    return a

class UnionFind:
    def __init__(self, x) -> None:
        self.uf = [-1] * x

    def find(self, x):
        r = x
        while self.uf[x] >= 0:
            x = self.uf[x]

        while r != x:
            self.uf[r], r = x, self.uf[r]
        return x

    def union(self, x, y):
        ux, uy = self.find(x), self.find(y)
        if ux == uy:
            return
        if self.uf[ux] >= self.uf[uy]:
            self.uf[uy] += self.uf[ux]
            self.uf[ux] = uy
        else:
            self.uf[ux] += self.uf[uy]
            self.uf[uy] = ux
        return

    def __print__(self):
        return self.uf

def spfa(x,g,n):
    dis = [float('inf') for i in range(n)]
    dis[x] = 0
    state = [False for i in range(n)]
    state[x] = True
    queue = collections.deque()
    queue.append(x)
    while queue:
        cur = queue.popleft()
        state[cur] = False
        for next_ in g[cur]:
            if dis[next_] > dis[cur] + 1:
                dis[next_] = dis[cur] + 1
                if state[next_] == False:
                    state[next_] = True
                    queue.append(next_)
    return dis

def gcd(x,y):
    if y == 0:
        return x
    return gcd(y,x % y)

def solve():
    n = int(input())
    a = rep()
    q = []
    prev = 0
    c = collections.Counter(a)
    ans = []
    for i in range(n + 1):
        ans.append(prev + c[i])
        if c[i] == 0:
            if not q:
                break
            else:
                q[-1][1] -= 1
                prev += i - q[-1][0]
                if q[-1][1] == 0:
                    q.pop()
        else:
            if c[i] > 1:
                q.append([i,c[i] - 1])
    ans.extend([-1 for i in range(n + 1 - len(ans))])
    print(*ans)




if __name__ == '__main__':
    for i in...