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 N = 1e5 + 5;
  const int mod = 998244353;
  int n;
  vector<int> dsk[N];
  long long f[N];
  int cur[N];
  bool ok = 1;
  void dfs(int u, int pre, int val) {
    if (((int)(dsk[u]).size()) == 1 && u != 1) {
      cur[u] = 1;
      return;
    }
    for (int v : dsk[u])
      if (v != pre) {
        dfs(v, u, val);
        cur[u] += cur[v];
      }
    if (cur[u] % val == 0)
      cur[u] = 1;
    else if (u != pre && (cur[u] + 1) % val == 0)
      cur[u] = 0;
    else
      ok = 0;
  }
  int solve(int val) {
    for (int i = 1; i <= n; ++i) cur[i] = 0;
    ok = 1;
    dfs(1, 1, val);
    return ok;
  }
  void gogo() {
    cin >> n;
    for (int i = 1; i <= n; ++i) dsk[i].clear();
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      dsk[u].push_back(v);
      dsk[v].push_back(u);
    }
    for (int i = 1; i <= n; ++i) f[i] = 0;
    f[1] = 1;
    for (int i = 1; i < n; ++i) (f[1] *= 2) %= mod;
    for (int i = 2; i <= n - 1; ++i)
      if ((n - 1) % i == 0) f[i] = solve(i);
    for (int i = n; i >= 1; --i)
      for (int j = 2 * i; j <= n; j += i) f[i] = (f[i] - f[j] + mod) % mod;
    for (int i = 1; i <= n; ++i) {
      if (i >= 2) assert(f[i] <= 1);
      cout << f[i] << ' ';
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    if (fopen("sol"
              ".inp",
              "r")) {
      freopen(
          "sol"
          ".inp",
          "r", stdin);
      freopen(
          "sol"
          ".out",
          "w", stdout);
    }
    int t;
    cin >> t;
    while (t--) gogo();
  }
  