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 = 300005;
  const int base = 998244353;
  int t, n, f[N], d[N], par[N];
  vector<int> a[N];
  void dfs(int u, int p, int c) {
    if (f[c] == 0) return;
    d[u] = 0;
    for (auto v : a[u])
      if (v != p) dfs(v, u, c);
    if (d[u] % c == 0)
      d[p]++;
    else if ((d[u] + 1) % c != 0)
      f[c] = 0;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    for (int iTest = 1; iTest <= t; iTest++) {
      cin >> n;
      for (int i = 1; i <= n; i++) a[i].clear(), f[i] = 0;
      for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        a[u].push_back(v);
        a[v].push_back(u);
      }
      f[n] = 0;
      f[1] = 1;
      for (int i = 1; i < n; i++) f[1] = (f[1] * 2) % base;
      for (int i = 2; i < n; i++)
        if ((n - 1) % i == 0) {
          f[i] = 1;
          dfs(1, 0, i);
        }
      for (int i = n - 1; i > 1; i--)
        for (int j = i * 2; j < n; j += i)
          if (f[j] > 0) {
            f[i] = 0;
            break;
          }
      for (int i = 2; i < n; i++) f[1] = (f[1] - f[i] + base) % base;
      for (int i = 1; i <= n; i++) cout << f[i] << " ";
      cout << "\n";
    }
    return 0;
  }
  