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;
  const int N = 1e5 + 10;
  const int P = 998244353;
  int T, n;
  int ans[N];
  vector<int> e[N];
  int f[N];
  bool dfs(int x, int fa, int y) {
    for (auto v : e[x])
      if (v != fa && !dfs(v, x, y)) return 0;
    if (f[x] % y == 0) {
      f[fa]--;
      return 1;
    }
    f[x]--;
    if (f[x] % y == 0) return 1;
    return 0;
  }
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cin >> T;
    while (T--) {
      cin >> n;
      for (int i = 1; i <= n; i++) e[i].clear();
      for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
      }
      if (n == 2) {
        cout << 2 << " " << 0 << "\n";
      } else {
        int rt = 0;
        for (int i = 1; i <= n; i++)
          if (((int)e[i].size()) > 1) rt = i;
        for (int i = 1; i <= n; i++) ans[i] = 0;
        for (int i = 1; i * i <= (n - 1); i++) {
          if ((n - 1) % i) continue;
          for (int j = 1; j <= n; j++) f[j] = ((int)e[j].size());
          if (dfs(rt, 0, i)) ans[i] = 1;
          for (int j = 1; j <= n; j++) f[j] = ((int)e[j].size());
          if (dfs(rt, 0, (n - 1) / i)) ans[(n - 1) / i] = 1;
        }
        for (int i = 2; i <= n; i++)
          for (int j = i + i; j <= n; j += i)
            if (ans[j]) ans[i] = 0;
        for (int i = 1; i < n; i++) ans[1] = 2 * ans[1] % P;
        for (int i = 2; i <= n; i++) ans[1] = (ans[1] - ans[i] + P) % P;
        for (int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
      }
    }
    return 0;
  }
  