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;
  using namespace chrono;
  template <typename T>
  T gcd(T a, T b) {
    return (b ? __gcd(a, b) : a);
  }
  template <typename T>
  T lcm(T a, T b) {
    return (a * (b / gcd(a, b)));
  }
  template <typename T, typename T1>
  T amax(T &a, T1 b) {
    if (b > a) a = b;
    return a;
  }
  template <typename T, typename T1>
  T amin(T &a, T1 b) {
    if (b < a) a = b;
    return a;
  }
  template <typename T1, typename T2>
  istream &operator>>(istream &in, pair<T1, T2> &a) {
    in >> a.first >> a.second;
    return in;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &out, pair<T1, T2> a) {
    out << "{" << a.first << ", " << a.second << "}";
    return out;
  }
  const int mod = 998244353;
  template <typename T>
  void print(T &a) {
    for (auto &x : a) cout << x << " ";
    cout << "\n";
  }
  int power(int x, int y) {
    int res = 1;
    while (y) {
      if (y & 1) res = res * 1ll * x % mod;
      y = y >> 1;
      x = x * 1ll * x % mod;
    }
    return res;
  }
  const int N = 1e5 + 10;
  vector<int> adj[N];
  void addedge(int u, int v) {
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  bool notpossible;
  int dfs(int v, int p, int d) {
    if (notpossible) {
      return 0;
    }
    int my = 0;
    for (auto &x : adj[v]) {
      if (x == p) continue;
      int tmp = dfs(x, v, d);
      if (tmp == 0)
        my++;
      else if (tmp + 1 == d)
        continue;
      else {
        notpossible = 1;
        return 0;
      }
    }
    return (my) % d;
  }
  int ans[N];
  int real_ans[N];
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < (n + 1); i++) adj[i].clear();
    for (int i = 0; i < (n - 1); i++) {
      int t1, t2;
      cin >> t1 >> t2;
      addedge(t1, t2);
    }
    ans[1] = power(2, (n - 1));
    for (int i = 2; i < (n + 1); i++) {
      if ((n - 1) % i) continue;
      notpossible = 0;
      int tmp = dfs(1, -1, i);
      if (notpossible || tmp) {
        ans[i] = 0;
      } else {
        ans[i] = 1;
      }
    }
    for (int i = 1; i < (n + 1); i++) {
      ;
    }
    for (int i = (n + 1) - 1; i >= 1; i--) {
      if ((n - 1) % i) {
        real_ans[i] = 0;
        continue;
      }
      int x = 2 * i;
      real_ans[i] = ans[i];
      while (x <= n) {
        if (i == 1) {
          ;
        }
        real_ans[i] -= real_ans[x];
        x += i;
      }
    }
    for (int i = 1; i < (n + 1); i++) cout << real_ans[i] << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    auto start1 = high_resolution_clock::now();
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    auto stop1 = high_resolution_clock::now();
    auto duration = duration_cast<milliseconds>(stop1 - start1);
    return 0;
  }
  