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;
  const long long MAX = 1e5 + 5;
  long long ans[MAX];
  vector<long long> P[MAX];
  long long pom[MAX];
  bool czy = true;
  long long akt;
  void dfs(long long u, long long fa) {
    for (auto it : P[u]) {
      if (it == fa) continue;
      dfs(it, u);
    }
    if (pom[u] % akt == 0) {
      if (fa != -1) pom[fa]++;
    } else {
      pom[u]++;
      if (pom[u] % akt != 0) czy = false;
    }
  }
  void solve() {
    long long n;
    cin >> n;
    for (long long i = 1; i <= n; i++) P[i].clear();
    for (long long i = 1; i <= n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      P[a].push_back(b);
      P[b].push_back(a);
    }
    ans[1] = 1;
    for (long long i = 1; i <= n - 1; i++) ans[1] = (ans[1] * 2) % mod;
    for (long long i = 2; i <= n; i++) {
      if ((n - 1) % i == 0) {
        czy = true;
        akt = i;
        for (long long j = 1; j <= n; j++) pom[j] = 0;
        dfs(1, -1);
        ans[i] = czy;
      } else
        ans[i] = 0;
    }
    for (long long i = n; i >= 1; i--) {
      for (long long j = 2 * i; j <= n; j += i)
        ans[i] = (ans[i] - ans[j] + mod) % mod;
    }
    for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    for (long long z = 0; z < t; z++) solve();
    return 0;
  }
  