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 = 100010, MOD = 998244353;
  int Q, n, tot, head[N], f[N];
  vector<int> d[N];
  struct edge {
    int next, to;
  } e[N * 2];
  void add(int from, int to) {
    e[++tot] = (edge){head[from], to};
    head[from] = tot;
  }
  int dfs(int x, int fa, int k) {
    int cnt = 0;
    for (int i = head[x]; ~i; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        int val = dfs(v, x, k);
        if (val == -1) return -1;
        cnt += val;
      }
    }
    if (cnt % k == 0) return 1;
    if (cnt % k == k - 1) return 0;
    return -1;
  }
  int main() {
    for (int i = 2; i < N; i++)
      for (int j = i; j < N; j += i) d[j].push_back(i);
    scanf("%d", &Q);
    while (Q--) {
      memset(head, -1, sizeof(head));
      memset(f, 0, sizeof(f));
      tot = 0;
      f[1] = 1;
      scanf("%d", &n);
      for (int i = 1, x, y; i < n; i++) {
        scanf("%d%d", &x, &y);
        add(x, y);
        add(y, x);
        f[1] = f[1] * 2 % MOD;
      }
      for (int i = 0; i < d[n - 1].size(); i++)
        if (dfs(1, 0, d[n - 1][i]) == 1) f[d[n - 1][i]] = 1;
      for (int i = n; i >= 1; i--)
        for (int j = i * 2; j <= n; j += i) f[i] -= f[j];
      for (int i = 1; i <= n; i++) cout << f[i] << " ";
      cout << "\n";
    }
    return 0;
  }
  