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 long long mod = 998244353;
  vector<int> G[100020];
  long long num[100010], dp[100010];
  int falg;
  long long qpow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = ans * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  void dfs(int x, int fa, int k) {
    if (falg == 1) return;
    for (int i = 0; i < G[x].size(); i++)
      if (G[x][i] != fa) dfs(G[x][i], x, k);
    if (dp[x] % k == 0)
      dp[fa]++;
    else {
      if (fa) dp[x]++;
      if (dp[x] % k != 0) falg = 1;
    }
    if (falg == 1) return;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n, a, b;
      cin >> n;
      for (int i = (0); i < (n + 1); ++i) G[i].clear(), num[i] = dp[i] = 0;
      for (int i = (0); i < (n - 1); ++i) {
        scanf("%d%d", &a, &b);
        G[a].push_back(b);
        G[b].push_back(a);
      }
      num[1] = qpow(2, n - 1);
      for (int i = (2); i < (n); ++i) {
        if ((n - 1) % i == 0) {
          falg = 0;
          dfs(1, 0, i);
          num[i] = 1 - falg;
          for (int i = (0); i < (n + 1); ++i) dp[i] = 0;
        }
      }
      for (int i = n; i >= 1; i--)
        for (int j = 2 * i; j <= n; j += i) num[i] -= num[j];
      for (int i = (1); i < (n + 1); ++i) cout << (num[i] + mod) % mod << ' ';
      cout << '\n';
    }
  }
  