description: You are given a tree with n nodes. As a reminder, a tree is a connected undirected graph without cycles.
  
  Let a_1, a_2, …, a_n be a sequence of integers. Perform the following operation exactly n times: 
  
    * Select an unerased node u. Assign a_u := number of unerased nodes adjacent to u. Then, erase the node u along with all edges that have it as an endpoint. 
  
  
  
  For each integer k from 1 to n, find the number, modulo 998 244 353, of different sequences a_1, a_2, …, a_n that satisfy the following conditions:
  
    * it is possible to obtain a by performing the aforementioned operations exactly n times in some order. 
    * \operatorname{gcd}(a_1, a_2, …, a_n) = k. Here, \operatorname{gcd} means the greatest common divisor of the elements in a. 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.
  
  The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).
  
  Each of the next n - 1 lines contains two integers u and v (1 ≤ u, v ≤ n) indicating there is an edge between vertices u and v. It is guaranteed that the given edges form a tree.
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
  
  Output
  
  For each test case, print n integers in a single line, where for each k from 1 to n, the k-th integer denotes the answer when \operatorname{gcd} equals to k.
  
  Example
  
  Input
  
  
  2
  3
  2 1
  1 3
  2
  1 2
  
  
  Output
  
  
  3 1 0
  2 0
  
  Note
  
  In the first test case, 
  
  <image>
  
    * If we delete the nodes in order 1 → 2 → 3 or 1 → 3 → 2, then the obtained sequence will be a = [2, 0, 0] which has \operatorname{gcd} equals to 2. 
    * If we delete the nodes in order 2 → 1 → 3, then the obtained sequence will be a = [1, 1, 0] which has \operatorname{gcd} equals to 1. 
    * If we delete the nodes in order 3 → 1 → 2, then the obtained sequence will be a = [1, 0, 1] which has \operatorname{gcd} equals to 1. 
    * If we delete the nodes in order 2 → 3 → 1 or 3 → 2 → 1, then the obtained sequence will be a = [0, 1, 1] which has \operatorname{gcd} equals to 1. 
  
  
  
  Note that here we are counting the number of different sequences, not the number of different orders of deleting nodes.
solution: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  using ull = unsigned long long;
  const int MONKE = 0;
  const ll mod = 998244353;
  void solve() {
    int n;
    cin >> n;
    vector<pair<int, int>> edges(n - 1);
    vector<vector<int>> adj(n);
    for (auto &[l, r] : edges) {
      cin >> l >> r;
      adj[l - 1].push_back(r - 1);
      adj[r - 1].push_back(l - 1);
    }
    vector<ll> f(n + 1);
    f[1] = 1;
    for (int i = 1; i < n; i++) f[1] = (f[1] + f[1]) % mod;
    function<int(int, int, int)> dfs = [&](int node, int par, int val) {
      int cnt = 0;
      for (auto e : adj[node]) {
        if (e == par) continue;
        int res = dfs(e, node, val);
        if (res == -1) return -1;
        cnt += res;
      }
      if (cnt % val == 0) return 1;
      if (cnt % val == val - 1) return 0;
      return -1;
    };
    for (int i = 2; i < n; i++) {
      if ((n - 1) % i == 0) {
        f[i] = (dfs(0, -1, i) > 0);
      }
    }
    for (int i = n; i > 0; i--) {
      for (int j = i + i; j < n; j += i) f[i] -= f[j];
      if (i == 1) f[i] = (f[i] + mod) % mod;
    }
    for (int i = 1; i <= n; i++) cout << f[i] << " \n"[i == n];
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    for (int tc = 1; tc <= t; tc++) {
      solve();
    }
    return MONKE;
  }
  