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:
// Problem: D. MEX Tree
// Contest: Codeforces - Codeforces Round #721 (Div. 2)
// URL: https://codeforces.com/contest/1527/problem/D
// Memory Limit: 256 MB
// Time Limit: 1500 ms

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (int)(x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

using namespace std;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
#define int long long
const int INF = 0x3f3f3f3f;
const double PI = acos(-1);
const int N = 2e5+10;

vector<int>e[N];
int ans[N],in[N],tim,out[N],vis[N],siz[N];

void dfs(int u,int f){
	in[u] = ++tim;
	siz[u] = 1;
	for(auto v:e[u]){
		if(v == f) continue;
		dfs(v,u);
		siz[u]+=siz[v];
	}
	out[u] = tim;
}

bool isson(int s,int f){
	return (in[f]<=in[s] && in[s]<=out[f]);
}

signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

	IOS;
	int T;
	cin>>T;
	while(T--){
		int n;
		cin>>n;
		tim = 0;
		for(int i=0;i<=n;i++) {
			e[i].clear();
			vis[i] = 0;
			ans[i] = 0;
		}
		for(int i=1;i<n;i++){
			int u,v;
			cin>>u>>v;
			e[u].pb(v);
			e[v].pb(u);
		}
		dfs(0,-1);
		ans[0] = ans[1] = n*(n-1)/2;
		int p = -1;
		for(auto v:e[0]) {
			ans[1] -= (siz[v]*(siz[v]-1)/2);
			if(isson(1,v)) p = v;
		}
		int l = 1 , r = -1;
		for(int i=1;i<n;i++){
			bool ok = false;
			if(isson(l,i)) ok = true;
			if(r!=-1 && isson(r,i)) ok = true;
			if(!ok){
				if(isson(i,l)) l = i;
				else if((r==-1 && !isson(i,p)) || isson(i,r) ) r = i;
				else break;
			}
			if(r == -1) ans[i+1] = siz[l]*(n-siz[p]);
			else ans[i+1] = siz[l]*siz[r];
		}
		for(int i=0;i<n;i++) ans[i]-=ans[i+1];
		for(int i=0;i<=n;i++) cout<<ans[i]<<" \n"[i==n];
		
		
		
	}





    return 0;
}