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 fx(l,n) inline l n
#define int long long
 
using std::vector;
 
const int N=1598244;
 
fx(int,gi)(){
	char c=getchar();int s=0,f=1;
	while(c>'9'||c<'0'){
		if(c=='-') f=-f;
		c=getchar();
	}
	while(c<='9'&&c>='0') s=(s<<3)+(s<<1)+(c-'0'),c=getchar();
	return s*f;
}
int T,dfnn,n,ans[N],pl,l=1,r=-1,u,v;
bool nos;
vector<int>e[N];
struct Tree{
	int tin,tout,siz;
	fx(void,clr)(){tin=tout=siz=0;}
}t[N];
fx(void,dfs)(int now,int fa){
	t[now].tin=dfnn++;
	for(int i=0;i<e[now].size();i++){
		if(e[now][i]==fa) continue;
		dfs(e[now][i],now);
	}
	t[now].tout=dfnn;
	t[now].siz=t[now].tout-t[now].tin;
//	printf("%lld <- %lld : [%lld %lld]\n",now,fa,t[now].t-in,t[now].tout);
}
signed main(){
	T=gi();
	while(T--){
		n=gi();dfnn=0;l=1,r=-1;pl=-1;
		for(int i=0;i<=n+100;i++) e[i].erase(e[i].begin(),e[i].end()),t[i].clr();
		memset(ans,0,sizeof(int)*(n+100));
		for(int i=1;i<n;i++){
			u=gi(),v=gi();
			e[u].push_back(v);
			e[v].push_back(u);
		}
		dfs(0,-1);
		ans[0]=ans[1]=n*(n-1)/2;
		for(int i=0;i<e[0].size();i++){
			ans[1]-=t[e[0][i]].siz*(t[e[0][i]].siz-1)/2;
			if(t[e[0][i]].tin<=t[1].tin&&t[e[0][i]].tout>t[1].tin) pl=e[0][i];
		}
//		printf("%lld %lld\n",ans[0],ans[1]);
		for(int i=1;i<n;i++){
			nos=0;
			if(t[i].tin<=t[l].tin&&t[i].tout>t[l].tin) nos=1;
			else if(~r&&t[i].tin<=t[r].tin&&t[i].tout>=t[r].tout) nos=1;
//			printf("pos: %lld %d\n",i,nos);
			if(!nos){
				if(t[l].tin<=t[i].tin&&t[i].tin<t[l].tout) l=i;
				else if((r==-1&&!(t[pl].tin<=t[i].tin&&t[i].tin<t[pl].tout))||(~r&&t[r].tin<=t[i].tin&&t[i].tin<t[r].tout)){
//					printf("%lld %lld\n",l,);
					r=i;
				}
				else break;
			}
//			printf("%lld %lld\n",i,r);
			ans[i+1]=t[l].siz*(r==-1?n-t[pl].siz:t[r].siz);
		}
		for(int i=0;i<=n;i++) printf("%lld ",ans[i]-ans[i+1]);
		printf("\n");
	}
}