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 long long mod = 998244353;
  const unsigned gen_seed =
      std::chrono::system_clock::now().time_since_epoch().count();
  std::mt19937_64 gen(gen_seed);
  std::vector<std::pair<int, int>> ord;
  vector<set<int>> g;
  std::vector<int> deg;
  void go(int v) {
    if (deg[v] == 1) {
      int to = *g[v].begin();
      ord.push_back(make_pair(v, to));
      g[to].erase(v);
      g[v].erase(to);
      deg[to]--;
      deg[v]--;
      go(to);
    }
  }
  int main() {
    int t;
    scanf("%d", &t);
    for (int afaf = 0; afaf < t; afaf++) {
      int n;
      scanf("%d", &n);
      g = vector<set<int>>(n);
      deg = std::vector<int>(n, 0);
      for (int i = 0; i < n - 1; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        u--;
        v--;
        g[u].insert(v);
        g[v].insert(u);
        deg[u]++;
        deg[v]++;
      }
      ord = std::vector<std::pair<int, int>>();
      for (int i = 0; i < n; i++) go(i);
      std::vector<int> ans(n + 1, 0);
      long long lft = 1;
      for (int i = 0; i < n - 1; i++) lft = lft * 2 % mod;
      for (int d = n - 1; d >= 2; d--) {
        if ((n - 1) % d == 0) {
          std::vector<int> val(n, 0);
          for (auto x : ord) {
            if (val[x.first] % d == 0) {
              val[x.second]++;
            } else
              val[x.first]++;
          }
          int can = 1;
          for (int i = 0; i < n; i++)
            if (val[i] % d != 0) can = 0;
          for (int d2 = 2 * d; d2 < n; d2 += d) {
            if (ans[d2] == 1) can = 0;
          }
          if (can == 1) {
            ans[d] = 1;
            lft = (lft + mod - 1) % mod;
          }
        }
      }
      ans[1] = lft;
      for (int i = 1; i < n + 1; i++) {
        printf("%d", ans[i]);
        if (i != n)
          printf(" ");
        else
          printf("\n");
      }
    }
  }
  