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 recur_depth = 0;
  const long long sz = 1e5 + 10, mod = 998244353;
  vector<long long> g[sz];
  long long ans[sz], dp[sz];
  void dfs(long long u, long long p, long long k) {
    for (long long& v : g[u]) {
      if (v == p) continue;
      dfs(v, u, k);
      if (dp[v] % k != 0)
        dp[v]++;
      else
        dp[u]++;
    }
  }
  int main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      scanf("%lld", &n);
      for (long long i = 1; i <= n - 1; ++i) {
        long long u, v;
        scanf("%lld", &u), scanf("%lld", &v);
        g[u].push_back(v);
        g[v].push_back(u);
      }
      long long m = n - 1;
      ans[1] = 1;
      for (long long i = 1; i <= m; ++i) ans[1] = (ans[1] * 2) % mod;
      for (long long i = 2; i <= m; i++) {
        if (m % i == 0) {
          for (long long j = 1; j <= n; ++j) dp[j] = 0;
          dfs(1, -1, i);
          bool ok = 1;
          for (long long j = 1; j <= n; ++j) ok &= (dp[j] % i == 0);
          ans[i] = ok;
        }
      }
      for (long long i = n; i >= 1; --i)
        for (long long j = i + i; j <= n; j += i) ans[i] -= ans[j];
      for (long long i = 1; i <= n; ++i) {
        if (ans[i] < 0) ans[i] += mod;
        printf("%lld ", ans[i]);
      }
      printf("\n");
      for (long long i = 1; i <= n; ++i) g[i].clear(), ans[i] = 0;
    }
    return 0;
  }
  