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 M = 998244353;
  int Tc, n, ans[100004], dp[100004];
  vector<int> g[100004];
  bool ok;
  void dfs(int x, int par, int C) {
    if (!ok) return;
    for (auto y : g[x])
      if (y != par) dfs(y, x, C);
    if (dp[x] % C != 0) {
      if (par) dp[x]++;
      if (dp[x] % C != 0) {
        ok = 0;
        return;
      }
    } else
      dp[par]++;
  }
  void solve() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) g[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      g[u].push_back(v), g[v].push_back(u);
    }
    for (int i = 1; i <= n; i++) ans[i] = 0;
    ans[1] = 1;
    for (int i = 1; i < n; i++) ans[1] = (ans[1] + ans[1]) % M;
    for (int i = 2; i <= n - 1; i++)
      if ((n - 1) % i == 0) {
        for (int j = 0; j <= n; j++) dp[j] = 0;
        ok = 1, dfs(1, 0, i), ans[i] = ok;
      }
    for (int i = n; i; i--)
      for (int j = i + i; j <= n; j += i) ans[i] = (ans[i] - ans[j] + M) % M;
    for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
    puts("");
  }
  int main() {
    for (scanf("%d", &Tc); Tc--;) solve();
  }
  