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 have(x,i) (st[x]<=st[i]&&ed[i]<=ed[x])
#define ll long long
using namespace std;
const int N=2e5+5;

int read()
{
	int s=0;
	char c=getchar(),lc='+';
	while (c<'0'||'9'<c) lc=c,c=getchar();
	while ('0'<=c&&c<='9') s=s*10+c-'0',c=getchar();
	return lc=='-'?-s:s;
}
void write(ll x)
{
	if (x<0) putchar('-'),x=-x;
	if (x<10) putchar(x+'0');
	else write(x/10),putchar(x%10+'0');
}
void print(ll x=-1,char c='\n')
{
	write(x);
	putchar(c);
}
struct edge
{
	int to,nxt;
}e[N*2];
int head[N],cnte=0;
void add_edge(int u,int v)
{
	e[++cnte].to=v;
	e[cnte].nxt=head[u];
	head[u]=cnte;
}
int st[N],ed[N],cnt,siz[N],bel[N],n;
void dfs(int now,int father)
{
	st[now]=++cnt;
	for (int i=head[now];i;i=e[i].nxt)
	{
		int to=e[i].to;
		if (to==father) continue;
		dfs(to,now);
	}
	ed[now]=cnt;
	siz[now]=ed[now]-st[now]+1;
}
void dfs1(int now,int father,int x)
{
	bel[now]=x;
	for (int i=head[now];i;i=e[i].nxt)
	{
		int to=e[i].to;
		if (to==father) continue;
		dfs1(to,now,x);
	}
}
void clear(int n)
{
	cnte=cnt=0;
	for (int i=0;i<n;i++) head[i]=0;
}
ll ans[N];
ll calc(int x,int y)
{
	if (!x) return 1LL*(n-siz[bel[y]])*siz[y];
	if (bel[x]==bel[y])
	{
		if (have(x,y)) return 1LL*(n-siz[bel[y]])*siz[y];
		if (have(y,x)) return 1LL*(n-siz[bel[x]])*siz[x];
		return 0;
	}
	return 1LL*siz[x]*siz[y];
}

signed main(signed Recall,char *_902_[])
{
	(void)Recall,(void)_902_;
	int T=read();
	while (T--)
	{
		n=read();
		clear(n);
		for (int i=1;i<n;i++)
		{
			int u=read(),v=read();
			add_edge(u,v);
			add_edge(v,u);
		}
		dfs(0,0);
		ans[0]=1LL*n*(n-1)/2;
		for (int i=1;i<=n+1;i++) ans[i]=0;
		int x=0,y=0,sum=1;
		for (int i=head[0];i;i=e[i].nxt)
		{
			int to=e[i].to;
			dfs1(to,0,to);
			ans[1]+=1LL*sum*siz[to];
			sum+=siz[to];
		}
		for (int i=1;i<n;i++)
		{
			if (!x&&!y) y=i;
			else
			{
				if (!have(i,y)&&!have(i,x))
				{
					if (have(y,i)) y=i;
					else if (have(x,i)) x=i;
					else break;
				}
				if (siz[x]<siz[y]) swap(x,y);
			}
			ans[i+1]=calc(x,y);
		}
		for (int i=0;i<=n;i++) print(ans[i]-ans[i+1],' ');puts("");
	}
	
	return 0;
}
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