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;
typedef long long LL;
const int NMAX = 2e5 + 5;

vector<int> graph[NMAX];
int sizeSubtree[NMAX], tata[NMAX], st, dr;
bool viz[NMAX];

void dfs1(int dad, int nod)
{
    tata[nod] = dad;
    sizeSubtree[nod] = 1;
    viz[nod] = 1;

    for(auto fiu : graph[nod])
        if(!viz[fiu])
        {
            dfs1(nod, fiu);

            sizeSubtree[nod] += sizeSubtree[fiu];
        }
}

int dfs2(int cautat, int nod)
{
    viz[nod] = 1;

    if(!viz[tata[nod]])
        return dfs2(cautat, tata[nod]);
    else
    {
        sizeSubtree[tata[nod]] -= sizeSubtree[nod];
        nod = tata[nod];

        if(nod == st)
            st = cautat;
        else if(nod == dr)
            dr = cautat;
        else
            return 0;

        return 1;
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int t;

    cin >> t;

    for(int test = 0; test < t; test++)
    {
        int n;

        cin >> n;

        for(int i = 0; i < n; i++)
        {
            graph[i].clear();
            viz[i] = 0;
        }

        for(int i = 0; i < n - 1; i++)
        {
            int a, b;

            cin >> a >> b;

            graph[a].push_back(b);
            graph[b].push_back(a);
        }

        dfs1(-1, 0);

        for(int i = 1; i < n; i++)
            viz[i] = 0;

        LL P = 1LL * n * (n - 1) / 2;
        LL suma = 0LL;

        for(auto fiu : graph[0])
            suma += 1LL * sizeSubtree[fiu] * (sizeSubtree[fiu] - 1) / 2;

        cout << suma << ' ';
        P -= suma;

        st = dr = 0;
        bool semafor = 0;

        for(int i = 1; !semafor && i < n; i++)
        {
            if(viz[i])
                cout << 0 << ' ';
            else if(!dfs2(i, i))
            {
                cout << P << ' ';

                for(int j = i + 1; j <= n; j++)
                    cout << 0 << ' ';

                i = n;
                semafor = 1;
            }
            else
            {
                cout << P - 1LL * sizeSubtree[st] * sizeSubtree[dr] << ' ';
                P = 1LL * sizeSubtree[st] * sizeSubtree[dr];
            }
        }

        if(!semafor)
            cout << 1;

        cout << '\n';
    }

    return 0;
}