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;
  void main0();
  int main() {
    clock_t start, end;
    ios::sync_with_stdio(false);
    cin.tie(0);
    main0();
    return 0;
  }
  const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
  const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
  const int N = 2e5 + 5;
  const int M = 1e5;
  const int INF = 0x3f3f3f3f;
  const long long INFF = 0x3f3f3f3f3f3f3f3f;
  const int mod = 998244353;
  const double eps = 1e-6;
  const double Pi = acos(-1.0);
  mt19937 rnd(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  long long qpow(long long n, long long m) {
    long long ret = 1;
    while (m) {
      if (m & 1) ret = ret * n % mod;
      n = n * n % mod;
      m >>= 1;
    }
    return ret;
  }
  vector<int> e[N];
  int ans[N];
  int succ;
  int dfs(int k, int u, int fa) {
    int cnt = 0;
    for (int v : e[u])
      if (v != fa) {
        cnt += dfs(k, v, u);
      }
    if (cnt % k == 0) return 1;
    if (cnt % k == k - 1) return 0;
    succ = 0;
    return 0;
  }
  void main0() {
    int T;
    cin >> T;
    while (T--) {
      int n;
      cin >> n;
      for (int i = 1; i <= n; ++i) {
        e[i].clear();
        ans[i] = (i == 1 ? qpow(2, n - 1) : 0);
      }
      for (int i = 1; i <= n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
      }
      for (int i = 2; i <= n - 1; ++i)
        if ((n - 1) % i == 0) {
          succ = 1;
          dfs(i, 1, -1);
          if (succ) ans[1]--, ans[i]++;
        }
      for (int i = 2; i <= n; ++i)
        if (ans[i]) {
          for (int j = 2; i * j <= n; ++j)
            if (ans[i * j]) {
              ans[1]++, ans[i]--;
              break;
            }
        }
      ans[1] += mod;
      ans[1] %= mod;
      for (int i = 1; i <= n; ++i) {
        cout << ans[i] << ' ';
      }
      cout << endl;
    }
  }
  