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 ll long long

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<vector<int>> g(n);
        vector<int> sub(n);
        vector<int> fa(n, -1);
        vector<bool> vis(n, false);
        vector<ll> mex(n + 2, 0);
        vis[0] = true;
        for (int i = 1; i < n; i++) {
            int u, v;
            cin >> u >> v;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        function<int(int, int)> dfs1 = [&](int now, int pre) {
            int res = 0;
            for (auto to : g[now]) {
                if (to == pre) continue;
                fa[to] = now;
                int subtree = dfs1(to, now);
                res += subtree;
                if (!now) {
                    mex[0] += 1LL * subtree * (subtree - 1) / 2;
                }
            }
            return sub[now] = res + 1;
        };
        dfs1(0, -1);
        cout << mex[0] << " ";
        mex[0] = 1LL * n * (n - 1) / 2 - mex[0];
        int l = 0, r = 0, fr = 0;
        auto cal = [&](int l, int r) {
            if (!r && !fr) {
                fr = 1;
                for (auto to : g[0]) {
                    if (vis[to]) {
                        sub[0] -= sub[to];
                    }
                }
            }
            return 1LL * sub[l] * sub[r];
        };
        for (int i = 1; i < n; i++) {
            if (vis[i]) {
                mex[i] = mex[i - 1];
            } else {
                int now = i;
                while (!vis[now]) {
                    vis[now] = true;
                    now = fa[now];
                }
                if (now == l) {
                    l = i;
                } else if (now == r) {
                    r = i;
                } else {
                    break;
                }
                mex[i] = cal(l, r);
            }
        }
        for (int i = 1; i <= n; i++) {
            cout << mex[i - 1] - mex[i] << " \n"[i == n];
        }
    }
    return 0;
}