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:
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define int long long
using namespace std;
//logic:https://codeforces.com/contest/1527/submission/116835971
//imp obs: the mex of path is i only if node with 0,1,2,,,,i-1 are there in single path
//so if we know the number of paths which have 0....i-1 then the number of paths which have from 0...i 
//can be easily found by finding if node i is within the paths which cover 0...i-1 or outside of the coverage
//look at code to understand better
const int N=200005,logn=18;
vector<int>g[N];
int depth[N],ans[N],sz[N],p[N][logn];
int n;
void dfs(int node,int par=-1)
{
    if(par==-1)
        depth[node]=0;
    else
        depth[node]=1+depth[par];
    sz[node]=1;
    p[node][0]=par;
    for(auto it:g[node])
    {
        if(it==par)
            continue;
        dfs(it,node);
        sz[node] += sz[it];
    }
}
int go_up_by_k(int node,int k)
{
    for(int j=logn-1;j>=0;j--)
        if((k>>j)&1)
            node=p[node][j];
    return node;
}
int lca(int x,int y)
{
    if(depth[x]>depth[y])
        swap(x,y);
    int k=(depth[y]-depth[x]);
    y=go_up_by_k(y,k);
    if(x==y)
        return x;
    for(int j=logn-1;j>=0;j--)
        if(p[x][j]!=p[y][j])
        {
            x=p[x][j];
            y=p[y][j];
        }
    return p[x][0];
}
int dist(int x,int y)
{
    int l=lca(x,y);
    int d=(depth[x]+depth[y]-(2*depth[l]));
    return d;
}
int nc2(int x)
{
    int res=(x*(x-1))/2;
    return res;
}
int calc(int x,int y)
{
    if(depth[x]>depth[y])
        swap(x,y);
    int l=lca(x,y);
    if(l==x)
    {
        int gotodepth=depth[y]-depth[x]-1;
        int node=go_up_by_k(y,gotodepth);
        return (sz[y]*(n-sz[node]));
    }
    else
    {
        return (sz[x]*sz[y]);
    }
}
void adjust(int& x,int& y,int nxt)
{
    if(dist(x,nxt)+dist(nxt,y) == dist(x,y))
        return;
    if(dist(x,nxt)+dist(x,y) == dist(nxt,y))
    {
        x=nxt;
        return;
    }
    if(dist(nxt,y)+dist(x,y) == dist(nxt,x))
    {
        y=nxt;
        return;
    }
    x=-1;
}
void solve()
{
    cin>>n;
    for(int j=0;j<logn;j++)
        for(int i=0;i<=n;i++)
            p[i][j]=-1;
    for(int i=1;i<n;i++)
    {
        int x,y;
        cin>>x>>y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs(0);
    for(int j=1;j<logn;j++)
    {
        for(int i=0;i<n;i++)
            p[i][j]=(p[i][j-1]==-1?-1:p[p[i][j-1]][j-1]);
    }
    int tot=nc2(n);
    int temp=0;
    for(auto it:g[0])
    {
        temp += nc2(sz[it]);
    }
    ans[0]=temp;
    tot -= temp;
    int x=0,y=1;
    for(int i=1;i<n;i++)
    {
        if(x==-1)
        {
            ans[i]=tot;
            tot=0;
            break;
        }
        int path=calc(x,y);
        ans[i]=tot-path;
        tot=path;
        if(i<(n-1))
        {
            adjust(x,y,i+1);
        }
    }
    ans[n]=tot;
    for(int i=0;i<=n;i++)
        cout<<ans[i]<<" ";
    cout<<"\n";
    for(int i=0;i<=n;i++)
    {
        ans[i]=0;
        g[i].clear();
    }
}
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t=1;
    cin>>t;
    while(t--)
        solve();
    return 0;
}