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 <iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
#define ls k<<1
#define rs k<<1|1
using namespace std;
typedef long long ll;
const int maxn=2e5+100;
int t,n;
vector<int> G[maxn];
int fa[maxn],siz[maxn],vis[maxn];
ll ans[maxn];
int st,en;
void dfs(int x,int f)
{
    siz[x]=1;
    fa[x]=f;
    for(int i=0;i<G[x].size();i++)
    {
        int to=G[x][i];
        if(to==f) continue;
        dfs(to,x);
        siz[x]+=siz[to];
    }
}
ll cal()
{
    if(st==en)
    {
        ll ans=0;
        for(int i=0;i<G[st].size();i++)
        {
            int to=G[st][i];
            ans+=siz[to]*2+1ll*(siz[to])*(n-1-siz[to]);//子树到根和子树到其他子树
        }
        return ans/2;
    }
    return 1ll*siz[st]*siz[en];
}
int add(int x)
{
    if(vis[x]) return 1;
    int which=x;
    int sum=siz[x];
    while(vis[which]==0)
    {
        vis[which]=1;
        siz[fa[which]]-=sum;
        sum+=siz[fa[which]];
        which=fa[which];
    }
    if(which==st)
    {
        st=x;
        return 1;
    }
    if(which==en)
    {
        en=x;
        return 1;
    }
    return 0;
}
int main()
{
  scanf("%d",&t);
  while(t--)
  {
      scanf("%d",&n);
      for(int i=0;i<=n+1;i++)
      {
          G[i].clear();
          fa[i]=-1;
          vis[i]=0;
          siz[i]=0;
          ans[i]=0;
      }
      for(int i=0,u,v;i<n-1;i++)
      {
          scanf("%d%d",&u,&v);
          G[u].push_back(v);
          G[v].push_back(u);
      }
      dfs(0,-1);
      st=en=0;
      ans[0]=1ll*n*(n-1)/2;
      vis[0]=1;
      vis[n]=1;
      for(int i=1;i<=n;i++)
      {
          ans[i]=cal();
          if(!add(i)) break;
      }
      for(int i=0;i<=n;i++)
        printf("%lld ",ans[i]-ans[i+1]);
      printf("\n");
  }
  return 0;
}