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> 
#define int long long
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define endl "\n"

using namespace std;

const int max_n = 2e5+100;
const int limit = 1e6+10;
const int max_k = 1e7+10;
const long long inf = 1e18;
const int off_set = 1e3+10;
const int Mod = 998244353;

int n;
vector<int> adj[max_n];
int ans[max_n],st[max_n],parent[max_n];
int pref[max_n];

void dfs(int v,int p){
	st[v]=1;
	parent[v]=p;

	for(int x:adj[v]) if(x!=p){
		dfs(x,v);
		st[v]+=st[x];
	}
}

signed main(){
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    int T;cin>>T;
    while(T--){
    	cin>>n;
    	for(int i=0;i<=n;i++) {
    		ans[i]=0;
    		pref[i]=0;
    		adj[i].clear();
    	}

    	for(int i=0;i<n-1;i++){
    		int x,y;cin>>x>>y;
    		adj[x].pb(y);
    		adj[y].pb(x);
    	}
    	dfs(0,-1);

    	ans[0]=(n*(n-1))/2;

    	int cur = 1;
    	for(int x:adj[0]){
    		ans[0]-=(st[x]*cur);
    		cur += st[x];
    	}

    	pref[0]=ans[0];
    	bool visited[n];
    	memset(visited,0,sizeof(visited));

    	int x=1,last=1;
    	while(x!=-1){
    		visited[x]=1;
    		if(x!=0) last=x;

    		x = parent[x];
    	}
    	ans[1]=((n*(n-1))/2)-ans[0];
    	ans[1]-=(n-st[last])*st[1];
    	pref[1]=pref[0]+ans[1];

    	int i=2;
    	pair<int,int> end_pts(0,1);
    	pair<int,int> subtrees(n-st[last],st[1]);

    	while(i<=n){
    		if(i==n){
    			ans[i]=((n*(n-1))/2)-pref[i-1];
    			break;
    		}

    		while(i<n && visited[i]) {
    			pref[i]=pref[i-1];
    			i++;
    		}

    		int x = i;
    		while(!visited[x]){
    			visited[x]=1;
    			x = parent[x];
    		}

    		if(x==end_pts.x){
    			ans[i]=((n*(n-1))/2)-pref[i-1]-(st[i]*subtrees.y);
    			pref[i]=pref[i-1]+ans[i];

    			end_pts.x = i;
    			subtrees.x = st[i];

    			i++;
    		}else if(x==end_pts.y){
    			ans[i]=((n*(n-1))/2)-pref[i-1]-(st[i]*subtrees.x);
    			pref[i]=pref[i-1]+ans[i];

    			end_pts.y = i;
    			subtrees.y = st[i];

    			i++;
    		}else{
    			ans[i]=((n*(n-1))/2)-pref[i-1];
    			break;
    		}
    	}

    	for(int i=0;i<=n;i++){
    		cout<<ans[i]<<" ";
    	}
    	cout<<endl;
    }
}