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;
  std::mt19937 rnd(
      (int)std::chrono::steady_clock::now().time_since_epoch().count());
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  const int MAXN = 100000;
  const int MOD = 998244353;
  int n;
  vector<int> adj[MAXN];
  int ans[MAXN + 1];
  int par[MAXN], sz[MAXN];
  void dfsinit(int at) {
    sz[at] = 1;
    for (int to : adj[at]) {
      if (to == par[at]) continue;
      par[to] = at;
      dfsinit(to);
      sz[at] += sz[to];
    }
  }
  void solve() {
    par[0] = -1;
    dfsinit(0);
    for (int i = (1); i <= (n); ++i) ans[i] = 0;
    ans[1] = 1;
    for (int i = (0); i < (n - 1); ++i) ans[1] = (long long)ans[1] * 2 % MOD;
    for (int i = (2); i <= (n); ++i)
      if ((n - 1) % i == 0) {
        bool ok = true;
        for (int j = (0); j < (n); ++j)
          if ((sz[j] - 1) % i != 0 && sz[j] % i != 0) ok = false;
        if (ok) ++ans[i];
      }
    for (int i = n; i >= 1; --i)
      for (int j = i + i; j <= n; j += i) ans[i] -= ans[j];
  }
  void run() {
    scanf("%d", &n);
    for (int i = (0); i < (n - 1); ++i) {
      int a, b;
      scanf("%d%d", &a, &b);
      --a, --b;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    solve();
    for (int i = (1); i <= (n); ++i) {
      if (i != 1) printf(" ");
      printf("%d", ans[i]);
    }
    puts("");
    for (int i = (0); i < (n); ++i) adj[i].clear();
  }
  int main() {
    int ncase;
    scanf("%d", &ncase);
    for (int i = (1); i <= (ncase); ++i) run();
    return 0;
  }
  