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 MAXN = 1e5 + 10, MOD = 998244353;
  int n;
  vector<int> adj[MAXN];
  bool work[MAXN], ans[MAXN];
  int dfs(int c, int p, int k) {
    int me = 0;
    for (auto nxt : adj[c])
      if (nxt != p) {
        int they = dfs(nxt, c, k);
        if (they == -1) return -1;
        assert(they == 0 || they == 1);
        me += they;
      }
    if ((me + 1) % k == 0) return 0;
    if (me % k == 0) return 1;
    return -1;
  }
  void solve() {
    cin >> n;
    for (int i = 0; i <= n; i++) adj[i].clear(), ans[i] = work[i] = 0;
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      cin >> a >> b, --a, --b;
      adj[a].push_back(b), adj[b].push_back(a);
    }
    for (int i = 2; i <= n; i++)
      if ((n - 1) % i == 0) {
        work[i] = dfs(0, -1, i) == 1;
      }
    int cur = 1;
    for (int i = 0; i < n - 1; i++) cur = cur * 2 % MOD;
    for (int i = 2; i <= n; i++) {
      ans[i] = work[i];
      for (int j = 2 * i; j <= n; j += i) ans[i] &= !work[j];
      cur -= ans[i];
    }
    for (int i = 1; i <= n; i++) {
      if (i == 1)
        cout << cur;
      else
        cout << ' ' << ans[i];
    }
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T = 1;
    cin >> T;
    while (T--) solve();
  }
  