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


def putin():
    return map(int, sys.stdin.readline().split())


def sol():
    n = int(sys.stdin.readline())
    G = []
    for i in range(n):
        G.append([])
    for i in range(n - 1):
        u, v = putin()
        G[u].append(v)
        G[v].append(u)
    parents = [0] * n
    parents[0] = -1
    capacity = [0] * n
    DFS_stack = [0]
    colors = [0] * n
    while DFS_stack:
        new_v = DFS_stack[-1]
        if colors[new_v] == 0:
            colors[new_v] = 1
            for elem in G[new_v]:
                if colors[elem] == 0:
                    DFS_stack.append(elem)
                    parents[elem] = new_v
        elif colors[new_v] == 1:
            colors[new_v] = 2
            DFS_stack.pop()
            S = 0
            for elem in G[new_v]:
                if colors[elem] == 2:
                    S += capacity[elem]
            capacity[new_v] = S + 1
        else:
            DFS_stack.pop()
    L = 0
    R = 0
    answer = [n * (n - 1) // 2]
    S = 0
    for children in G[0]:
        S += capacity[children] * (capacity[children] - 1) // 2
    answer.append(n * (n - 1) // 2 - S)
    cur_v = 0
    path = {0}
    while len(answer) < n + 1:
        cur_v += 1
        ancestor = cur_v
        while ancestor not in path:
            path.add(ancestor)
            if parents[ancestor] == 0:
                capacity[0] -= capacity[ancestor]
            ancestor = parents[ancestor]
        if ancestor != cur_v:
            if ancestor != L and ancestor != R:
                break
            if ancestor == L:
                L = cur_v
            else:
                R = cur_v
        answer.append((capacity[L]) * (capacity[R]))
    while len(answer) < n + 2:
        answer.append(0)
    for i in range(n + 1):
        print(answer[i] - answer[i + 1], end=' ')
    print()


for iter in range(int(sys.stdin.readline())):
    sol()