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;
  long long t;
  vector<long long> adj[100001];
  long long ss[100001];
  void dfs(long long no, long long par = -1) {
    ss[no] = 1;
    for (auto j : adj[no]) {
      if (j != par) {
        dfs(j, no);
        ss[no] += ss[j];
      }
    }
  }
  const long long mod = 998244353;
  long long ans[100001];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      for (long long i = 0; i < n; i++) {
        adj[i].clear();
      }
      for (long long i = 0; i < n - 1; i++) {
        long long aa, bb;
        cin >> aa >> bb;
        aa--;
        bb--;
        adj[aa].push_back(bb);
        adj[bb].push_back(aa);
      }
      dfs(0);
      vector<pair<long long, long long>> ss2;
      vector<long long> tt;
      for (long long i = 1; i <= n; i++) {
        ans[i] = 0;
      }
      for (long long i = 1; i < n; i++) {
        ss2.push_back({ss[i] - 1, ss[i]});
      }
      for (long long i = 1; i <= n - 1; i++) {
        if ((n - 1) % i == 0) {
          tt.push_back(i);
        }
      }
      reverse(tt.begin(), tt.end());
      for (auto j : tt) {
        ans[j] = 1;
        for (auto k : ss2) {
          long long co = 0;
          if ((k.first) % j == 0) {
            co++;
          }
          if ((k.second) % j == 0) {
            co++;
          }
          ans[j] = (ans[j] * co) % mod;
        }
        for (auto i : tt) {
          if (i > j) {
            if ((i % j) == 0) {
              ans[j] = (ans[j] - ans[i] + mod) % mod;
            }
          }
        }
      }
      for (long long i = 1; i <= n; i++) {
        cout << ans[i] << " ";
      }
      cout << '\n';
    }
    return 0;
  }
  