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 N = 1e5 + 50;
  const long long mod = 998244353;
  vector<long long> g[N];
  long long F[N];
  long long H[N];
  long long binpow(long long a, long long b) {
    long long ret = 1;
    while (b) {
      if (b & 1) {
        ret = (ret * a) % mod;
      }
      a = (a * a) % mod;
      b >>= 1;
    }
    return ret;
  }
  void reset(long long n) {
    for (long long i = 1; i <= n; i++) {
      g[i].clear();
      F[i] = 0;
    }
  }
  long long getF(long long node, long long k, long long par = -1) {
    long long val = 0;
    for (auto to : g[node]) {
      if (to == par) continue;
      long long res = getF(to, k, node);
      if (res == -1) {
        return -1;
      }
      if (res == 0) val++;
    }
    if (val % k == 0) {
      return 0;
    } else if (val % k == k - 1) {
      return 1;
    } else {
      return -1;
    }
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      for (long long i = 0; i < n - 1; i++) {
        long long u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
      }
      F[1] = binpow(2, n - 1);
      for (long long i = 2; i <= n; i++) {
        if ((n - 1) % i == 0) {
          if (getF(1, i) != -1) F[i] = 1;
        }
      }
      for (long long i = n; i >= 1; i--) {
        H[i] = F[i];
        for (long long j = i * 2; j <= n; j += i) {
          H[i] -= H[j];
        }
      }
      for (long long i = 1; i <= n; i++) cout << H[i] << ' ';
      cout << '\n';
      reset(n);
    }
  }
  