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 target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
// #include<ext/pb_ds/assoc_container.hpp>
// #include<ext/pb_ds/tree_policy.hpp>
// #include<ext/pb_ds/tag_and_trait.hpp>
// using namespace __gnu_pbds;
// #include<boost/multiprecision/cpp_int.hpp>
// namespace multiprecisioninteger = boost::multiprecision;
// using cint=multiprecisioninteger::cpp_int;
using namespace std;
using ll=long long;
using datas=pair<ll,ll>;
using ddatas=pair<long double,long double>;
using tdata=pair<ll,datas>;
using vec=vector<ll>;
using mat=vector<vec>;
using pvec=vector<datas>;
using pmat=vector<pvec>;
// using llset=tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;
#define For(i,a,b) for(i=a;i<(ll)b;++i)
#define bFor(i,b,a) for(i=b,--i;i>=(ll)a;--i)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)
#define brep(i,N) bFor(i,N,0)
#define brep1(i,N) bFor(i,N,1)
#define all(v) (v).begin(),(v).end()
#define allr(v) (v).rbegin(),(v).rend()
#define vsort(v) sort(all(v))
#define vrsort(v) sort(allr(v))
#define uniq(v) vsort(v);(v).erase(unique(all(v)),(v).end())
#define endl "\n"
#define popcount __builtin_popcountll
#define eb emplace_back
#define print(x) cout<<x<<endl
#define printyes print("Yes")
#define printno print("No")
#define printYES print("YES")
#define printNO print("NO")
#define output(v) do{bool f=0;for(auto outi:v){cout<<(f?" ":"")<<outi;f=1;}cout<<endl;}while(0)
#define matoutput(v) do{for(auto outimat:v)output(outimat);}while(0)
constexpr ll mod=1000000007;
// constexpr ll mod=998244353;
constexpr ll inf=1LL<<60;
constexpr long double eps=1e-9;
const long double PI=acosl(-1);
template<class T,class E> ostream& operator<<(ostream& os,const pair<T,E>& p){return os<<"("<<p.first<<","<<p.second<<")";}
template<class T> ostream& operator<<(ostream& os,const vector<T>& v){
  os<<"{";bool f=false;
  for(auto& x:v){if(f)os<<",";os<<x;f=true;}
  os<<"}";
  return os;
}
template<class T> ostream& operator<<(ostream& os,const set<T>& v){
  os<<"{";bool f=false;
  for(auto& x:v){if(f)os<<",";os<<x;f=true;}
  os<<"}";
  return os;
}
template<class T> ostream& operator<<(ostream& os,const multiset<T>& v){
  os<<"{";bool f=false;
  for(auto& x:v){if(f)os<<",";os<<x;f=true;}
  os<<"}";
  return os;
}
template<class T,class E> ostream& operator<<(ostream& os,const map<T,E>& v){
  os<<"{";bool f=false;
  for(auto& x:v){if(f)os<<",";os<<x;f=true;}
  os<<"}";
  return os;
}
template<class T> inline bool chmax(T& a,const T b){bool x=a<b;if(x)a=b;return x;}
template<class T> inline bool chmin(T& a,const T b){bool x=a>b;if(x)a=b;return x;}
#ifdef DEBUG
void debugg(){cout<<endl;}
template<class T,class... Args>void debugg(const T& x,const Args&... args){cout<<" "<<x;debugg(args...);}
#define debug(...) cout<<__LINE__<<" ["<<#__VA_ARGS__<<"]:",debugg(__VA_ARGS__)
#else
#define debug(...) (void(0))
#endif

inline void startupcpp(void) noexcept{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(15);
}

ll modinv(ll a,const ll m=mod) noexcept{
  ll b=m,u=1,v=0,t;
  while(b){
    t=a/b;
    a-=t*b; swap(a,b);
    u-=t*v; swap(u,v);
  }
  return (u+m)%m;
}

ll moddevide(const ll a,const ll b) noexcept{return (a*modinv(b))%mod;}

vec modncrlistp,modncrlistm;

ll modncr(const ll n,const ll r) noexcept{
  if(n<r)return 0;
  ll i,size=modncrlistp.size();
  if(size<=n){
    modncrlistp.resize(n+1);
    modncrlistm.resize(n+1);
    if(!size){
      modncrlistp[0]=modncrlistm[0]=1;
      size++;
    }
    For(i,size,n+1)modncrlistp[i]=modncrlistp[i-1]*i%mod;
    modncrlistm[n]=modinv(modncrlistp[n]);
    for(i=n;i>size;--i)modncrlistm[i-1]=modncrlistm[i]*i%mod;
  }
  return modncrlistp[n]*modncrlistm[r]%mod*modncrlistm[n-r]%mod;
}

ll modpow(ll a,ll n,const ll m=mod){
  if(n<0)return 0;
  ll res=1;
  while(n>0){
    if(n&1)res=res*a%m;
    a=a*a%m;
    n>>=1;
  }
  return res;
}

constexpr ll gcd(const ll a,const ll b) noexcept{return (!b)?abs(a):(a%b==0)?abs(b):gcd(b,a%b);}
constexpr ll lcm(const ll a,const ll b) noexcept{return a/gcd(a,b)*b;}

ll N;
mat tree;
vec par,sz;
ll dfs(ll now,ll p=-1){
  par[now]=p;
  sz[now]=1;
  for(auto x:tree[now]){
    if(x==p)continue;
    sz[now]+=dfs(x,now);
  }
  return sz[now];
}
int main(){
  startupcpp();
int codeforces;cin>>codeforces;while(codeforces--){
  ll i;
  cin>>N;
  par.resize(N);
  sz.resize(N);
  tree.assign(N,vec());
  rep(i,N-1){
    ll a,b;
    cin>>a>>b;
    tree[a].emplace_back(b);
    tree[b].emplace_back(a);
  }
  dfs(0);
  mat g(N);
  vec vis(N,0),cnt(N+2,0);
  cnt[0]=cnt[1]=N*(N-1)/2;
  for(auto x:tree[0])cnt[1]-=sz[x]*(sz[x]-1)/2;
  vis[0]=1;
  ll L=0,R=0;
  rep1(i,N){
    cnt[i+1]=cnt[i];
    if(vis[i])continue;
    ll pos=i;
    while(!vis[pos]){
      vis[pos]=1;
      g[i].emplace_back(pos);
      pos=par[pos];
    }
    reverse(all(g[i]));
    if(pos!=L&&pos!=R){
      cnt[i+1]=0;
      break;
    }
    if(pos==L){
      L=i;
      cnt[i+1]=R?sz[R]*sz[L]:(N-sz[g[1][0]])*sz[L];
    }else{
      R=i;
      cnt[i+1]=L?sz[R]*sz[L]:sz[R]*(N-sz[g[1][0]]);
    }
  }
  vec ans(N+1);
  rep(i,N+1)ans[i]=cnt[i]-cnt[i+1];
  output(ans);
}
}