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 MAX = 1e5 + 7;
  const int MOD = 998244353;
  int t;
  int n;
  vector<int> G[MAX];
  int f[MAX], k;
  bool dfs(int u, int fa) {
    f[u] = 0;
    if (u != 1 && G[u].size() == 1) return true;
    for (auto v : G[u]) {
      if (v == fa) continue;
      if (dfs(v, u)) f[u]++;
    }
    if (fa && f[u] % k != 0) {
      f[u]++;
      return false;
    }
    return true;
  }
  int qw[MAX];
  int ans[MAX];
  int main() {
    ios::sync_with_stdio(false), cin.tie(NULL);
    qw[0] = 1;
    for (int i = 1; i <= 100000; i++) qw[i] = qw[i - 1] * 2 % MOD;
    cin >> t;
    while (t--) {
      cin >> n;
      for (int i = 1; i <= n; i++) G[i].clear(), ans[i] = 0;
      for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v), G[v].push_back(u);
      }
      ans[1] = qw[n - 1];
      for (int i = 2; i < n; i++) {
        if ((n - 1) % i != 0) continue;
        k = i;
        dfs(1, 0);
        int fg = 0;
        for (int j = 1; j <= n; j++)
          if (f[j] % i) {
            fg = 1;
            break;
          }
        if (!fg) ans[i]++;
      }
      for (int i = n; i >= 1; i--) {
        for (int j = 2 * i; j <= n; j += i) ans[i] -= ans[j];
      }
      for (int i = 1; i <= n; i++) cout << ans[i] << " ";
      cout << '\n';
    }
    return 0;
  }
  