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;
  inline int read() {
    int f = 1, ans = 0;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      ans = ans * 10 + c - '0';
      c = getchar();
    }
    return f * ans;
  }
  const int MAXN = 3e5 + 11;
  vector<int> vec[MAXN];
  int cas, N, Ans[MAXN], nw[MAXN];
  bool ff;
  void dfs(int u, int fath, int Lim) {
    nw[u] = 0;
    for (auto v : vec[u])
      if (v != fath) dfs(v, u, Lim);
    if (fath) {
      if (nw[u])
        nw[u] = (nw[u] + 1) % Lim;
      else
        nw[fath] = (nw[fath] + 1) % Lim;
    }
    if (nw[u]) ff = 0;
    return;
  }
  signed main() {
    cas = read();
    while (cas--) {
      N = read();
      for (int i = 1; i <= N; i++) vec[i].clear(), Ans[i] = 0;
      int res = 1;
      for (int i = 1; i < N; i++) {
        res = res * 2 % 998244353;
        int u = read(), v = read();
        vec[u].push_back(v), vec[v].push_back(u);
      }
      int cnt = 0;
      for (int i = 2; i <= N; i++)
        if (!((N - 1) % i)) {
          ff = 1;
          dfs(1, 0, i);
          Ans[i] = ff;
        }
      for (int i = N - 1; i >= 2; i--)
        for (int j = i + i; j <= N - 1; j += i) Ans[i] -= Ans[j];
      for (int i = 2; i <= N - 1; i++) cnt += Ans[i];
      Ans[1] = (res - cnt + 998244353) % 998244353;
      for (int i = 1; i <= N; i++) printf("%d ", Ans[i]);
      printf("\n");
    }
    return 0;
  }
  