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 + 9, mod = 998244353;
  vector<int> g[N];
  int dp[N], d, ok, ans[N];
  void dfs(int u, int p = 0) {
    if (!ok) return;
    for (auto v : g[u]) {
      if (v ^ p) {
        dfs(v, u);
      }
    }
    if (dp[u] % d != 0) {
      if (p) {
        dp[u]++;
      }
      if (dp[u] % d != 0) {
        ok = 0;
        return;
      }
    } else {
      dp[p]++;
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
      }
      for (int i = 1; i <= n; i++) {
        ans[i] = 0;
      }
      ans[1] = 1;
      for (int k = 1; k <= n - 1; k++) {
        ans[1] = (ans[1] + ans[1]) % mod;
      }
      for (d = 2; d <= n - 1; d++) {
        if ((n - 1) % d == 0) {
          ok = 1;
          dfs(1);
          ans[d] = ok;
          for (int i = 0; i <= n; i++) {
            dp[i] = 0;
          }
        }
      }
      for (int i = n; i >= 1; i--) {
        for (int j = i + i; j <= n; j += i) {
          ans[i] = (ans[i] - ans[j] + mod) % mod;
        }
      }
      for (int i = 1; i <= n; i++) {
        if (i > 1) cout << ' ';
        cout << ans[i];
      }
      cout << '\n';
      for (int i = 1; i <= n; i++) {
        g[i].clear();
      }
    }
    return 0;
  }
  