#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
const ll mod = 998244353;
int n;
int h[N],e[2*N],ne[2*N],idx,f[N],c[N];
ll size[N],ans[N];
void add(int a,int b)
{
	e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
void dfs(int u) 
{
	size[u]=1;
	for(int i=h[u];i!=-1;i=ne[i])
	{
		int v=e[i];
		if(f[u]==v) continue;
		f[v]=u;
		dfs(v);
		size[u]+=size[v];
	}
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		memset(h,-1,sizeof(h));
		idx=0;
		scanf("%d",&n);
		int a,b;
		for(int i=1;i<n;i++)
		{
			scanf("%d%d",&a,&b);
			add(a,b),add(b,a);//连边 
		}
		dfs(0);//以 0为根，求得子树的顶点数 size 
		ans[0]=1LL*n*(n-1)/2;//所有路径 
		ll sum=1;
		for(int i=h[0];i!=-1;i=ne[i])
		{
			int v=e[i];
			ans[1]+=sum*size[v];//ans[1]是包含0的所有路径 
			sum+=size[v];
		}
		a=0,b=0,c[0]=1;
		int p,x;
		for(int i=1;i<n;i++)
		{
			if(!c[i])
			{
				x=p=i;
				while(!c[x]) c[x]=1,p=x,x=f[x];
				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");
		for(int i=0;i<=n;i++)
		{
			size[i]=ans[i]=f[i]=c[i]=0;
		}
	} 
	return 0;
} 
### Prompt

Please formulate a Cpp solution to the following 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. 

### Response

```cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
const ll mod = 998244353;
int n;
int h[N],e[2*N],ne[2*N],idx,f[N],c[N];
ll size[N],ans[N];
void add(int a,int b)
{
	e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
void dfs(int u) 
{
	size[u]=1;
	for(int i=h[u];i!=-1;i=ne[i])
	{
		int v=e[i];
		if(f[u]==v) continue;
		f[v]=u;
		dfs(v);
		size[u]+=size[v];
	}
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		memset(h,-1,sizeof(h));
		idx=0;
		scanf("%d",&n);
		int a,b;
		for(int i=1;i<n;i++)
		{
			scanf("%d%d",&a,&b);
			add(a,b),add(b,a);//连边 
		}
		dfs(0);//以 0为根，求得子树的顶点数 size 
		ans[0]=1LL*n*(n-1)/2;//所有路径 
		ll sum=1;
		for(int i=h[0];i!=-1;i=ne[i])
		{
			int v=e[i];
			ans[1]+=sum*size[v];//ans[1]是包含0的所有路径 
			sum+=size[v];
		}
		a=0,b=0,c[0]=1;
		int p,x;
		for(int i=1;i<n;i++)
		{
			if(!c[i])
			{
				x=p=i;
				while(!c[x]) c[x]=1,p=x,x=f[x];
				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");
		for(int i=0;i<=n;i++)
		{
			size[i]=ans[i]=f[i]=c[i]=0;
		}
	} 
	return 0;
} 
```