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;

#define pb push_back
#define F first
#define S second
#define debug(x) cour<<#x<<" :"<<x<<"\n"
#define all(x) x.begin(),x.end()
#define pii pair<int,int>
#define FAST ios_base::sync_with_stdio(false), cin.tie(), cout.tie();
#define int long long

typedef long long ll;
typedef long double ld;

const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const int mlog = 20;
const int SQ = 400;

int par[maxn], sz[maxn], mark[maxn];
vector<int> adj[maxn];

int n;

void dfs(int v)
{
	sz[v] = 1;
	for(int u : adj[v]) if(u != par[v]) par[u] = v, dfs(u), sz[v] += sz[u];
}

int dp[maxn];

int visit(int v)
{
	while(!mark[v])
	{
		if(par[v] == 0) sz[0] -= sz[v];

		mark[v] = true;
		v = par[v];
	}

	return v;
}

void clear()
{
	for(int i = 0; i <= n; i++)
	{
		adj[i].clear();
		mark[i] = false;
		dp[i] = 0;
		par[i] = 0;
		sz[i] = 0;
	}
}

int32_t main()
{
    FAST;

    int T; cin >> T;
    while(T--)
    {
		cin >> n;

		for(int i = 1; i < n; i++)
		{
			int u, v;
			cin >> u >> v;

			adj[u].pb(v), adj[v].pb(u);
		}

		par[0] = -1;
		dfs(0);

		// base

		dp[0] = dp[1] = n * (n - 1) / 2;
		for(int u : adj[0]) dp[1] -= sz[u] * (sz[u] - 1) / 2;
		
		int L, R; L = R = 0;
		mark[0] = true;

		for(int i = 2; i <= n; i++)
		{
			if(mark[i - 1]) { dp[i] = dp[i - 1]; continue; }

			int v = visit(i - 1);

			if(v == L) L = i - 1;
			else if(v == R) R = i - 1;
			else break;

			dp[i] = sz[L] * sz[R];
		}

		for(int i = 0; i <= n; i++) 
			dp[i] -= dp[i + 1];


		for(int i = 0; i <= n; i++) cout << dp[i] << " ";
		cout << "\n";

		clear();
    }

    return 0;
}