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 collections import deque
input = sys.stdin.readline

def solve():
    n = int(input())
    G = [[] for _ in range(n)]
    for _ in range(n-1):
        a,b = map(int,input().split())
        G[a].append(b)
        G[b].append(a)

    size = [0]*n
    par = [-1]*n

    stk = [0]
    visited = [False]*n

    while stk:
        x = stk[-1]
        if not visited[x]:
            visited[x] = True
            for y in G[x]:
                if not visited[y]:
                    par[y] = x
                    stk.append(y)
        else:
            stk.pop()
            for y in G[x]:
                size[x] += size[y]
            size[x] += 1

    visited = [False]*n
    ans = [0]*(n+1)
    for y in G[0]:
        ans[0] += size[y] * (size[y]-1) // 2
    
    P = n*(n-1)//2
    P -= ans[0]

    visited[0] = True

    l,r = 0,0
    for i in range(1,n):
        if visited[i]:
            ans[i] = P - size[r] * size[l]
            continue

        u = i
        acc = 0
        while not visited[u]:
            visited[u] = True
            if par[u] == 0:
                size[0] -= size[u]
            u = par[u]

        if u == l:
            ans[i] = P - size[r] * size[i]
            l = i
            P = size[r] * size[l]
        elif u == r:
            ans[i] = P - size[l] * size[i]
            r = i
            P = size[r] * size[l]
        else:
            ans[i] = P
            P = 0
            break

    ans[-1] = P 
    print(*ans)
        
for nt in range(int(input())):
    solve()