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 long long mod = 998244353;
  bool dfs(vector<int> adj[], int p, int c, int a[], int k, bool &isposs) {
    a[c] = 0;
    for (auto u : adj[c])
      if (u != p) {
        bool f = dfs(adj, c, u, a, k, isposs);
        if (!f) {
          a[c]++;
        }
      }
    if (a[c] % k && p != -1) {
      a[c]++;
      if (a[c] % k) {
        isposs = false;
      }
      return true;
    }
    if (p == -1 && a[c] % k) {
      isposs = false;
    }
    return false;
  }
  long long fastexp(long long n, long long k) {
    if (k == 0)
      return 1;
    else if (k % 2)
      return (fastexp(n, k - 1) * n) % mod;
    else {
      long long w = fastexp(n, k / 2);
      return (w * w) % mod;
    }
  }
  vector<int> facts[100001];
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    for (int i = 2; i <= 100000; i++) {
      for (int j = 2 * i; j <= 100000; j += i) {
        facts[j].push_back(i);
      }
    }
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      vector<int> adj[n];
      int a[n];
      for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
      }
      int ans[n + 1];
      for (int i = 1; i <= n; i++) {
        ans[i] = 0;
      }
      ans[1] = fastexp(2, n - 1);
      for (auto f : facts[n - 1]) {
        bool ths = true;
        dfs(adj, -1, 0, a, f, ths);
        ans[f] = ths;
      }
      if (n != 2) {
        bool ths = true;
        dfs(adj, -1, 0, a, n - 1, ths);
        ans[n - 1] = ths;
      }
      for (int j = n - 1; j >= 2; j--) {
        if (ans[j]) {
          for (auto f : facts[j]) {
            ans[f] -= ans[j];
          }
          ans[1] -= ans[j];
        }
      }
      ans[1] += mod;
      ans[1] %= mod;
      for (int i = 1; i <= n; i++) {
        cout << ans[i] << " ";
      }
      cout << "\n";
    }
    return 0;
  }
  