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<cstdio>
#include<iostream>
#define N 1000017
#define ll long long
using namespace std;

struct edge{
	int v,next;
}e[N];
int tot,head[N],father[N];
ll P[N],siz[N],n;//Pi表示 mex>i的路径大小 
bool vis[N];

void add(int u,int v)
{
	e[++tot].v = v;
	e[tot].next = head[u];
	head[u] = tot;
}

void dfs1(int u,int fa)
{
	siz[u] = 1;
	father[u] = fa;
	for(int i = head[u];i;i = e[i].next)
	{
		int v = e[i].v;
		if(v == fa)continue;
		dfs1(v,u);
		siz[u] += siz[v];
	}
}

void clear()
{
	tot = 0;
	for(int i = 0;i < n;i += 1)
		vis[i] = siz[i] = head[i] = P[i] = 0;
}

int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lld",&n);
		for(int i = 1,u,v;i < n;i += 1){
			scanf("%d%d",&u,&v);
			add(u,v);
			add(v,u);
		}
		dfs1(0,-1);//计算子树size
		//计算res表示mex = 0答案 
		ll res = 0;
		for(int i = head[0];i;i = e[i].next){
			int v = e[i].v;
			res += siz[v] * (siz[v]-1)/2;
		}
		//计算P1 
		P[0] = (n-1)*n/2 - res;
		int cur = 1;
		vis[0] = vis[1] = 1;
		while(father[cur] != 0){
			cur = father[cur];
			vis[cur] = 1;
		}
		int nxt = cur;
		P[1] = (siz[0] - siz[cur]) * siz[1];
		int l = 0;
		int r = 1;
		for(int i = 2;i < n;i += 1)
		{
			if(vis[i])P[i] = P[i-1];
			else {
				cur = i;
				while(!vis[cur])cur = father[cur];
				//	cout << cur;
				if(cur != l && cur != r){
					P[i] = 0;
					break;
				}
				if(cur == l)P[i] = siz[r] * siz[i],l = i;
				else if(cur == r){
					if(l == 0)P[i] = (siz[l] - siz[nxt]) * siz[i];
					else P[i] = siz[l] * siz[i];
					r = i;
				}
				cur = i;
				while(!vis[cur]){
					vis[cur] = 1;
					cur = father[cur];
				}
			}
		}
		for(int i = 0;i <= n;i += 1)
		{
			//cout << "P" << P[i] << ' ';
			if(i == 0)printf("%lld ",res);
			else printf("%lld ",P[i-1] - P[i]);
		}
		clear();
		printf("\n");
	}
	return 0;
}