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 JAVA):

Solution:
// upsolve with kaiboy
import java.io.*;
import java.util.*;

public class CF1527D extends PrintWriter {
	CF1527D() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1527D o = new CF1527D(); o.main(); o.flush();
	}

	int[] sz, pp, eo; int[][] ej;
	void append(int i, int j) {
		int o = eo[i]++;
		if (o >= 2 && (o & o - 1) == 0)
			ej[i] = Arrays.copyOf(ej[i], o << 1);
		ej[i][o] = j;
	}
	int dfs(int p, int i) {
		pp[i] = p;
		int k = 1;
		for (int o = eo[i]; o-- > 0; ) {
			int j = ej[i][o];
			if (j != p)
				k += dfs(i, j);
		}
		return sz[i] = k;
	}
	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			sz = new int[n]; pp = new int[n]; eo = new int[n]; ej = new int[n][2];
			for (int h = 0; h < n - 1; h++) {
				int i = sc.nextInt();
				int j = sc.nextInt();
				append(i, j);
				append(j, i);
			}
			dfs(-1, 0);
			int u = 1, v = 0;
			boolean[] used = new boolean[n];
			used[u] = used[v] = true;
			while (pp[u] != 0) {
				u = pp[u];
				used[u] = true;
			}
			long[] aa = new long[n];
			for (int o = eo[0]; o-- > 0; ) {
				int j = ej[0][o];
				aa[0] += (long) sz[j] * (sz[j] - 1) / 2;
			}
			int k = sz[0] - sz[u];
			for (int o = eo[0]; o-- > 0; ) {
				int j = ej[0][o];
				if (j != u)
					aa[1] += (long) (k - 1 - sz[j]) * sz[j];
			}
			aa[1] /= 2;
			aa[1] += (long) (sz[u] - sz[1]) * (k - 1);
			aa[1] += sz[0] - sz[1] - 1;
			long m = (long) n * (n - 1) / 2 - aa[0] - aa[1];
			u = 1;
			for (int i = 2; i < n; i++) {
				if (used[i])
					continue;
				int w = i;
				while (!used[w]) {
					used[w] = true;
					w = pp[w];
				}
				if (w != u && w != v) {
					aa[i] = m; m = 0;
					break;
				}
				if (w == u) {
					aa[i] = (long) (sz[u] - sz[i]) * k;
					u = i;
				} else {
					aa[i] = (long) sz[u] * (k - sz[i]); k = sz[i];
					v = i;
				}
				m -= aa[i];
			}
			for (int i = 0; i < n; i++)
				print(aa[i] + " ");
			print(m);
			println();
		}
	}
}