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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  long long int gcd(long long int A, long long int B) {
    if (!B) return A;
    return gcd(B, A % B);
  }
  vector<vector<int>> adj(100005);
  vector<int> dp(100005, 0);
  bool dfs(int v, int u, int k) {
    int cur = 0;
    for (int x : adj[v]) {
      if (x == u) continue;
      cur += dfs(x, v, k);
    }
    bool res = 0;
    if (cur % k && u != -1) {
      cur++;
      res = 1;
    }
    if (cur % k) dp[k] = 0;
    return res ^ 1;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      for (int i = 0; i < n; i++) {
        adj[i].clear();
        dp[i + 1] = 0;
      }
      for (int i = 0; i < n - 1; i++) {
        int v, u;
        cin >> v >> u;
        v--;
        u--;
        adj[v].push_back(u);
        adj[u].push_back(v);
      }
      n--;
      vector<int> fac = {1};
      dp[1] = 1;
      for (int i = 0; i < n; i++) {
        dp[1] *= 2;
        if (dp[1] >= (long long int)(998244353))
          dp[1] -= (long long int)(998244353);
      }
      for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
          dp[i] = 1;
          dfs(0, -1, i);
          fac.push_back(i);
          if (i * i != n) {
            dp[n / i] = 1;
            dfs(0, -1, n / i);
            fac.push_back(n / i);
          }
        }
      }
      if (n != 1) {
        dp[n] = 1;
        fac.push_back(n);
        dfs(0, -1, n);
      }
      sort(fac.begin(), fac.end());
      for (int i = (int)(fac.size()) - 1; i >= 0; i--) {
        for (int j = 2 * fac[i]; j <= n; j += fac[i]) {
          dp[fac[i]] -= dp[j];
        }
      }
      dp[1] += (long long int)(998244353);
      if (dp[1] >= (long long int)(998244353))
        dp[1] -= (long long int)(998244353);
      for (int i = 1; i <= n + 1; i++) cout << dp[i] << " ";
      cout << '\n';
    }
  }
  