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("O3")
  const double PI = acos(-1);
  template <class A, class B>
  std::ostream& operator<<(std::ostream& st, const std::pair<A, B> p) {
    st << "(" << p.first << ", " << p.second << ")";
    return st;
  }
  void tc();
  auto test_cases = []() {
    int T;
    scanf("%d", &T);
    while (T--) tc();
  };
  using namespace std;
  const int mod = 998244353;
  int add(int a, int b) {
    a += b;
    if (a >= mod) a -= mod;
    if (a < 0) a += mod;
    return a;
  }
  int mul(int a, int b) { return (a * (long long)b) % mod; }
  vector<std::vector<int> > G;
  int n;
  std::pair<int, int> dfs(int u, int p, int m) {
    int ctr = 0;
    for (int v : G[u])
      if (v != p) {
        auto [c, e] = dfs(v, u, m);
        if (!c) return std::pair<int, int>(0, 0);
        ctr += 1 - e;
      }
    if (ctr % m == 0) return std::pair<int, int>(1, 0);
    if ((ctr + 1) % m == 0) return std::pair<int, int>(1, 1);
    return std::pair<int, int>(0, 0);
  }
  int solve(int k) {
    if ((n - 1) % k) return 0;
    if (k == 1) {
      int ret = 1;
      for (int i = 0; i < (n - 1); i++) ret = mul(ret, 2);
      return ret;
    }
    return dfs(1, 0, k).first;
  }
  int g[1 << 20];
  void tc() {
    scanf("%d", &n);
    G.clear();
    G.resize(n + 1);
    for (int e = 0; e < (n - 1); e++) {
      int a, b;
      scanf("%d %d", &a, &b);
      G[a].push_back(b);
      G[b].push_back(a);
    }
    for (int k = n; k >= 1; k--) {
      g[k] = solve(k);
      for (int s = 2 * k; s <= n; s += k) g[k] = add(g[k], -g[s]);
    }
    for (int k = 1; k <= n; k++) printf("%d%c", g[k], " \n"[k == n]);
  }
  int main() {
    test_cases();
    return 0;
  }
  