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;
  vector<long long> a;
  bool valido;
  vector<vector<long long>> adj;
  long long d;
  void dfs(long long u, long long p = 0) {
    if (!valido) return;
    for (long long v : adj[u])
      if (v != p) dfs(v, u);
    if (a[u] % d == 0) {
      a[p]++;
    } else {
      a[u]++;
      if (p == 0 || a[u] % d != 0) valido = 0;
    }
  }
  int32_t main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      a.resize(n + 1);
      adj = vector<vector<long long>>(n + 1);
      for (long long i = 0; i < n - 1; i++) {
        long long a, b;
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
      }
      vector<long long> freq(n + 1);
      freq[1] = 2;
      for (long long i = 1; i < n - 1; i++) freq[1] = 2 * freq[1] % 998244353;
      for (d = 2; d <= n - 1; d++) {
        if ((n - 1) % d == 0) {
          valido = 1;
          fill(a.begin(), a.end(), 0);
          dfs(1);
          freq[d] = valido;
        }
      }
      vector<long long> res(n + 1);
      for (long long d = n; d >= 1; d--) {
        res[d] = freq[d];
        for (long long mult = 2 * d; mult <= n; mult += d) {
          res[d] = res[d] - res[mult] + 998244353;
          res[d] %= 998244353;
        }
      }
      for (long long d = 1; d <= n; d++) {
        cout << res[d] << ' ';
      }
      cout << endl;
    }
    return 0;
  }
  