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<debug.h>
#include<bits/stdc++.h>
#define int long long
#define fi first
#define se second
#define pb push_back
using namespace std;

const int N = 2e5 + 10, MOD = 1e9+7, inf = 0x3f3f3f3f, INF = 0x3f3f3f3f3f3f3f3fll;

signed main()
{
	ios::sync_with_stdio(false);
	int T;
	cin >> T;
	while (T--) {
		int n;
		cin >> n;
		vector<vector<int> > G(n);
		for (int i = 0; i < n - 1; i++) {
			int u, v;
			cin >> u >> v;
			G[u].pb(v);
			G[v].pb(u);
		}
		vector<int> ans(n + 1);
		int tim = 0;
		vector<int> in(n), out(n), siz(n);
		function<void(int, int)> dfs = [&](int u, int fa) {
			in[u] = tim++;
			for (auto v : G[u]) if (v != fa) {
				dfs(v, u);
			}
			out[u] = tim;
			siz[u] = out[u] - in[u];
		};
		dfs(0, -1);
		ans[0] = ans[1] = n * (n - 1) / 2;
		int l = -1;
		for (auto v : G[0]) {
			ans[1] -= siz[v] * (siz[v] - 1) / 2;
			if (in[v] <= in[1] && in[1] < out[v]) {//1的父亲是v 
				l = v;
			}
		}
		int u = 1, v = -1;
		for (int i = 1; i < n; i++) {
			bool ok = false;
			if (in[i] <= in[u] && in[u] < out[i]) {//u的父亲是i 
				ok = true;
			}
			if (v != -1 && in[i] <= in[v] && in[v] < out[i]) {//v的父亲是i 
				ok = true; 
			} 
			if (!ok) { //u和v的父亲都不是i时 
				if (in[u] <= in[i] && in[i] < out[u]) {//i的父亲是u时 
					u = i;
				} else if ((v == -1 && !(in[l] <= in[i] && in[i] < out[l])) || (in[v] <= in[i] && in[i] < out[v])) {
					v = i;	//如果i的父亲是l或者i的父亲是v 
				} else {
					break;
				}
			}
			if (v == -1) {
				ans[i + 1] = siz[u] * (n - siz[l]);
			}
			else {
				ans[i + 1] = siz[u] * siz[v];
			}
		}
		for (int i = 0; i < n; i++) {
			ans[i] -= ans[i + 1];
		}
		for (int i = 0; i <= n; i++) {
			cout << ans[i] << " \n"[i == n];
		}
	} 
    return 0;
}