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 int N = 200020, MOD = 998244353;
  inline int read() {
    int s = 0, w = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
      if (ch == '-') w = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * w;
  }
  int T, n, u, v, f[N];
  bool flag;
  vector<int> e[N];
  int dfs(int k, int fa, int gr) {
    int cnt = 0;
    for (int i = 0; i < e[k].size(); i++) {
      int to = e[k][i];
      if (to == fa) continue;
      cnt = cnt + dfs(to, k, gr);
    }
    if (cnt % gr == 0) return 1;
    if ((cnt + 1) % gr == 0) return 0;
    flag = 0;
    return 0;
  }
  int main() {
    T = read();
    while (T--) {
      n = read();
      for (int i = 1; i <= n; i++) e[i].clear();
      memset(f, 0, sizeof(f)), f[1] = 1;
      for (int i = 1; i <= n - 1; i++) {
        f[1] = f[1] * 2 % MOD;
        u = read(), v = read();
        e[u].push_back(v), e[v].push_back(u);
      }
      for (int i = 2; i <= n - 1; i++)
        if ((n - 1) % i == 0) {
          flag = 1;
          if (dfs(1, 0, i) == 1 && flag) f[i] = 1;
        }
      for (int i = n - 1; i >= 1; i--)
        for (int j = 2; i * j <= n - 1; j++) f[i] = f[i] - f[i * j];
      f[1] = (f[1] + MOD) % MOD;
      for (int i = 1; i <= n; i++) printf("%d ", f[i]);
      printf("\n");
    }
    return 0;
  }
  