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;
  int n;
  vector<int> adjList[300005];
  vector<int> visited(300005);
  vector<int> a(300005);
  vector<long long> f(300005);
  vector<long long> twoPowers(300005);
  long long mod = 998244353;
  vector<long long> ans(300005);
  int dfs(int node, int goal) {
    visited[node] = 1;
    int accrued = 0;
    for (int i = 0; i < adjList[node].size(); i++) {
      int child = adjList[node][i];
      if (!visited[child]) {
        int au = dfs(child, goal);
        if (au == -1) return -1;
        accrued += au;
      }
    }
    int toReturn = 1;
    if (accrued % goal == 0)
      toReturn = 1;
    else if ((accrued + 1) % goal == 0)
      toReturn = 0;
    else
      toReturn = -1;
    return toReturn;
  }
  int main() {
    iostream::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    twoPowers[0] = 1;
    for (int i = 1; i < 300005; i++) {
      twoPowers[i] = (twoPowers[i - 1] * 2) % mod;
    }
    int t;
    cin >> t;
    while (t--) {
      cin >> n;
      for (int i = 0; i < n + 1; i++) {
        vector<int> v;
        adjList[i] = v;
        f[i] = 0;
        ans[i] = 0;
      }
      for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        adjList[u].push_back(v);
        adjList[v].push_back(u);
      }
      f[1] = twoPowers[n - 1];
      for (int i = 2; i <= n - 1; i++) {
        if ((n - 1) % i == 0) {
          for (int j = 0; j <= n; j++) {
            visited[j] = 0;
            a[j] = 0;
          }
          int res = dfs(0, i);
          if (res == 1) f[i] = 1;
        }
      }
      for (int i = n - 1; i >= 1; i--) {
        ans[i] = f[i];
        for (int j = 2; j * i <= n - 1; j++) ans[i] = ans[i] - ans[j * i];
      }
      for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
      cout << endl;
    }
  }
  