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>
using namespace std;

const int maxn=200010;

int T,n;
int fa[maxn],c[maxn],tot;
int ver[maxn<<1],nxt[maxn<<1],head[maxn];
long long ans[maxn],size[maxn];

template<typename Tp>
void read(Tp &x)
{
	x=0;int f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}x*=f;
}

void add(int u,int v)
{
	ver[++tot]=v;
	nxt[tot]=head[u];
	head[u]=tot;
}

void dfs(int x)
{
	size[x]=1;
	for(int i=head[x];i;i=nxt[i])
	{
		int y=ver[i];
		if(y==fa[x]) continue;
		fa[y]=x;
		dfs(y);
		size[x]+=size[y];
	}
}

int main()
{
	read(T);
	while(T--)
	{
		read(n);
		for(int i=1;i<n;i++) 
		{
			int u,v;
			read(u),read(v);
			add(u,v),add(v,u);
		}
		dfs(0);
		ans[0]=(long long)n*(n-1)/2;
		int sum=1;
		for(int i=head[0];i;i=nxt[i]) 
		{
			int y=ver[i];
			ans[1]+=size[y]*sum;
			sum+=size[y];
		}
		c[0]=1;
		int a=0,b=0;
 		for(int i=1;i<n;i++)
		{
			if(!c[i])
			{
				int x=i,p=x;
				while(!c[x]) c[x]=1,p=x,x=fa[x];
	//			cout<<x<<" "<<i<<endl;
				if(x==a)
				{
					size[a]-=size[p];
					a=i;
				}
				else if(x==b)
				{
//					size[b]-=size[p];
					b=i;
				}
				else break;
			}
			ans[i+1]=size[a]*size[b];
		}
		for(int i=0;i<=n;i++) printf("%lld ",ans[i]-ans[i+1]);
		printf("\n");
		tot=0;
		for(int i=0;i<=n+1;i++) head[i]=ans[i]=size[i]=c[i]=0;
	}
	return 0;
}