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:
from sys import stdin, gettrace

if gettrace():
    def inputi():
        return input()
else:
    def input():
        return next(stdin)[:-1]


    def inputi():
        return stdin.buffer.readline()


def readGraph(n, m):
    adj = [[] for _ in range(n)]
    for _ in range(m):
        u, v = map(int, input().split())
        adj[u].append(v)
        adj[v].append(u)
    return adj


def readTree(n):
    return readGraph(n, n - 1)


def euler_dfs(n, adj):
    parent = [None] * n
    depth = [-1] * n
    depth[0] = 0
    stack = [(0, adj[0])]
    euler = []
    eulerfirst = [-1] * n
    eulerlast = [-1] * n
    dcount = [0] * n
    pcount = [0] * n
    seen = 0

    while stack:
        p, aa = stack.pop()
        if eulerfirst[p] == -1:
            pcount[p] = seen
            seen += 1
            eulerfirst[p] = len(euler)
        eulerlast[p] = len(euler)
        dcount[p] = seen - pcount[p]
        euler.append(p)
        for c in aa[::-1]:
            if c != parent[p]:
                parent[c] = p
                depth[c] = depth[p] + 1
                stack.append((p, []))
                stack.append((c, adj[c]))
    return eulerfirst, eulerlast, parent, depth, dcount


def solve():
    n = int(input())
    adj = readTree(n)
    eulerfirst, eulerlast, parent, depth, dcount = euler_dfs(n, adj)

    def is_decendant(u, v):
        return eulerfirst[v] < eulerfirst[u] < eulerlast[v]

    res = [0] * (n + 1)
    paths = [0] * n
    res[0] = sum(dcount[v] * (dcount[v] - 1) for v in adj[0]) // 2
    paths[0] = (n * (n - 1)) // 2 - res[0]
    oneb = 1
    while oneb not in adj[0]:
        oneb = parent[oneb]
    paths[1] = dcount[1] * (n - dcount[oneb])
    pends = [1]
    for i in range(2, n):
        for x, e in enumerate(pends):
            if is_decendant(e, i):
                break
            elif is_decendant(i, e):
                pends[x] = i
                break
        else:
            if len(pends) == 2:
                break
            p = i
            q = pends[0]
            while depth[p] > depth[q]:
                p = parent[p]
            while depth[q] > depth[p]:
                q = parent[q]
            while p != q:
                p = parent[p]
                q = parent[q]
            if p != 0:
                break
            pends.append(i)
        if len(pends) == 2:
            paths[i] = dcount[pends[0]] * dcount[pends[1]]
        else:
            paths[i] = dcount[pends[0]] * (n - dcount[oneb])
    for i in range(1, n):
        res[i] = paths[i - 1] - paths[i]
    res[n] = paths[n - 1]
    print(' '.join(map(str, res)))


def main():
    t = int(input())
    for _ in range(t):
        solve()


if __name__ == "__main__":
    main()