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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std ;
int main()
{
    std::ios::sync_with_stdio(false) , cin.tie(0) ;
    int T ;
    cin >> T ;
    while(T --)
    {
        int n ;
        cin >> n ;
        vector<vector<int>> g(n + 1) ;
        for(int i = 1 ; i <= n - 1 ; i ++)
        {
            int u , v ;
            cin >> u >> v ;
            g[u].push_back(v) ;
            g[v].push_back(u) ;
        }
        vector<int> f(n + 1 , 0) ;
        vector<int> siz(n + 1 , 0) ;
        vector<int> siz2(n + 1 , 0) ;
        function<void(int , int)> dfs = [&](int fa , int u)
        {
            siz[u] = 1 ;
            f[u] = fa ;
            for(auto v : g[u])
            {
                if(v == fa)  continue ;
                dfs(u , v) ;
                siz[u] += siz[v] ;
            }
        } ;
        dfs(0 , 0) ;
        function<void(int , int , int)> dfs2 = [&](int fa , int u , int num)
        {
            siz2[u] = num ;
            for(auto v : g[u])
            {
                if(v == fa)  continue ;
                dfs2(u , v , num) ;
            }
        } ;
        for(auto v : g[0])  dfs2(0 , v , siz[v]) ;
        //ans[i]��ʾ����[0 , i - 1]��·����
        vector<long long> ans(n + 2 , 0) ;
        ans[0] = 1ll * n * (n - 1) / 2 ;
        ans[1] = ans[0] ;
        auto cal = [&](int x)
        {
            return 1ll * x * (x - 1) / 2 ;
        } ;
        for(auto v : g[0])  ans[1] -= cal(siz[v]) ;
        bool flag = true ;
        int l = 0 , r = 0 ;
        vector<bool> vis(n + 1 , false) ;
        vis[0] = true ;
        for(int i = 1 ; i <= n - 1 ; i ++)
        {
            int t = i ;
            if(!vis[t])
            {
                while(!vis[t])  vis[t] = true , t = f[t] ;
                if(!flag)  ans[i + 1] = 0 ;
                else
                {
                    if(t == l)  l = i ;
                    else if(t == r)  r = i ;
                    else  flag = false ;
                }
            }
            if(!flag)  ans[i + 1] = 0 ;
            else
            {
                if(r == 0)  ans[i + 1] = 1ll * siz[l] * (n - siz2[l]) ;
                else  ans[i + 1] = 1ll * siz[l] * siz[r] ;
            }
        }
        for(int i = 0 ; i <= n - 1 ; i ++)  ans[i] -= ans[i + 1] ;
        for(int i = 0 ; i <= n ; i ++)  cout << ans[i] << " \n"[i == n] ;
    }
    return 0 ;
}