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;
  long long PosMod(long long a) { return a >= 0 ? a : a + 998244353; }
  long long AddMod(long long a, long long b) {
    return PosMod((a + b) % 998244353);
  }
  long long MultMod(long long a, long long b) {
    return PosMod((a * b) % 998244353);
  }
  long long powll(long long a, long long n) {
    long long r = 1;
    for (long long b = n; b > 0; b >>= 1, a = MultMod(a, a))
      if (b & 1) r = MultMod(r, a);
    return r;
  }
  vector<vector<int>> adj;
  bool fail;
  int dfs(int here, int p, int targetK) {
    if (fail) return 0;
    int incoming = 0;
    for (int there : adj[here]) {
      if (there == p) continue;
      incoming += dfs(there, here, targetK);
    }
    if (incoming % targetK == 0)
      return 1;
    else if ((incoming + 1) % targetK == 0 && here != 1)
      return 0;
    else {
      fail = true;
      return 0;
    }
  }
  int n;
  void solve() {
    cin >> n;
    adj = vector<vector<int>>(n + 1);
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      adj[v].push_back(u);
      adj[u].push_back(v);
    }
    vector<long long> f = vector<long long>(n + 1);
    f[1] = powll(2, (long long)n - 1);
    for (int k = 2; k <= n - 1; k++) {
      if ((n - 1) % k != 0) continue;
      fail = false;
      dfs(1, 0, k);
      f[k] = fail ? 0 : 1;
    }
    vector<long long> h = vector<long long>(n + 1);
    h[1] = f[1];
    for (int i = 2; i <= n; i++) {
      long long val = f[i];
      if (val > 0) {
        for (int k = 2; i * k <= n; k++) {
          if (f[i * k]) val = 0;
        }
      }
      h[i] = val;
      h[1] -= val;
    }
    for (int i = 1; i <= n; i++) {
      printf("%lld ", h[i]);
    }
    cout << '\n';
  }
  int main() {
    int t;
    cin >> t;
    while (t--) solve();
  }
  