Problem: You are given a tree with n nodes, numerated from 0 to n-1. For each k between 0 and n, inclusive, you have to count the number of unordered pairs (u,v), u ≠ v, such that the MEX of all the node labels in the shortest path from u to v (including end points) is k.

The MEX of a sequence of integers is the smallest non-negative integer that does not belong to the sequence.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. 

The first line of each test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^{5}).

The next n-1 lines of each test case describe the tree that has to be constructed. These lines contain two integers u and v (0 ≤ u,v ≤ n-1) denoting an edge between u and v (u ≠ v).

It is guaranteed that the given edges form a tree.

It is also guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^{5}.

Output

For each test case, print n+1 integers: the number of paths in the tree, such that the MEX of all the node labels in that path is k for each k from 0 to n.

Example

Input


2
4
0 1
0 2
2 3
2
1 0


Output


1 2 1 1 1 
0 0 1 

Note

  1. In example case 1, <image>
    * For k = 0, there is 1 path that is from 2 to 3 as MEX([2, 3]) = 0. 
    * For k = 1, there are 2 paths that is from 0 to 2 as MEX([0, 2]) = 1 and 0 to 3 as MEX([0, 2, 3]) = 1. 
    * For k = 2, there is 1 path that is from 0 to 1 as MEX([0, 1]) = 2. 
    * For k = 3, there is 1 path that is from 1 to 2 as MEX([1, 0, 2]) = 3 
    * For k = 4, there is 1 path that is from 1 to 3 as MEX([1, 0, 2, 3]) = 4. 
  2. In example case 2, <image>
    * For k = 0, there are no such paths. 
    * For k = 1, there are no such paths. 
    * For k = 2, there is 1 path that is from 0 to 1 as MEX([0, 1]) = 2. 

Solution (in PYTHON3):

Solution:

import sys
from sys import stdin

from collections import deque
def NC_Dij(lis,start):

    N = len(lis)
    ret = [float("inf")] * len(lis)
    ret[start] = 0
    chnum = [1] * N
    
    q = deque([start])
    plis = [i for i in range(len(lis))]

    while len(q) > 0:
        now = q.popleft()

        for nex in lis[now]:

            if ret[nex] > ret[now] + 1:
                ret[nex] = ret[now] + 1
                plis[nex] = now
                q.append(nex)

    td = [(ret[i],i) for i in range(N)]
    td.sort()
    td.reverse()

    for tmp,i in td:
        if plis[i] != i:
            chnum[ plis[i] ] += chnum[i]

    return ret,plis,chnum

tt = int(stdin.readline())

for loop in range(tt):

    n = int(stdin.readline())

    lis = [ [] for i in range(n) ]
    for i in range(n-1):
        u,v = map(int,stdin.readline().split())
        lis[u].append(v)
        lis[v].append(u)

    
    dlis,plis,chnum = NC_Dij(lis,0)
    ans = [0] * (n+1)
    ans[0] = n * (n-1)//2

    in_flag = [False] * n
    in_flag[0] = True
    l,r = 0,0
    llast = None #parent is 0 and l is child

    flag = True
    for v in range(n):

        nv = v

        if in_flag[nv] == False:
            while True:
                if in_flag[nv] == True:
                    if nv == l:
                        l = v
                    elif nv == r:
                        r = v
                    else:
                        flag = False
                    break
                
                in_flag[nv] = True
                nv = plis[nv]

        if not flag:
            break

        if l == 0 and r == 0:
            nans = n * (n-1)//2
            for nex in lis[0]:
                nans -= chnum[nex] * (chnum[nex]-1)//2
        elif r == 0:
            if llast == None:
                tv = l
                while plis[tv] != 0:
                    tv = plis[tv]
                llast = tv

            nans = chnum[l] * (n-chnum[llast])

        else:
            nans = chnum[l] * chnum[r]

        ans[v+1] = nans
    
    for i in range(n):
        ans[i] -= ans[i+1]
    

    #print (chnum)
    print (*ans)