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>
  #pragma GCC optimize("-O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  mt19937_64 rnd(time(0));
  long long MOD = 998244353;
  const long double E = 1e-10;
  long long n, m, k, T, ans[500005];
  vector<long long> c[500005];
  long long binpow(long long h, long long r, long long md) {
    long long l = 1;
    while (r) {
      if (r & 1) l = (l * h) % md;
      h = (h * h) % md;
      r /= 2;
    }
    return l;
  }
  long long read() {
    char c = getchar();
    while (!isdigit(c)) {
      c = getchar();
    }
    long long res = 0;
    while (isdigit(c)) {
      res = res * 10 + c - '0';
      c = getchar();
    }
    return res;
  }
  long long gcd(long long x, long long y) {
    while (x && y) {
      if (x < y) swap(x, y);
      x %= y;
    }
    return x + y;
  }
  pair<long long, long long> dfs(int h, int p = -1) {
    long long z = 0;
    for (int i = 0; i < c[h].size(); i++)
      if (c[h][i] != p) {
        pair<long long, long long> u = dfs(c[h][i], h);
        if (!u.first) return make_pair(0, 0);
        z += u.second;
      }
    if (z % T == 0)
      return make_pair(1, 1);
    else if ((z + 1) % T == 0)
      return make_pair(1, 0);
    else
      return make_pair(0, 0);
  }
  void solve() {
    cin >> n;
    long long x = n - 1, p = sqrt(x);
    vector<long long> v;
    for (int i = 1; i <= n; i++) ans[i] = 0, c[i].clear();
    for (int i = 1; i <= p; i++)
      if (x % i == 0) {
        v.push_back(i);
        if (x / i != i) v.push_back(x / i);
      }
    for (int i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      c[x].push_back(y);
      c[y].push_back(x);
    }
    ans[1] = binpow(2, n - 1, MOD);
    for (int i = 1; i < v.size(); i++) {
      T = v[i];
      pair<long long, long long> o = dfs(1);
      if (o.first && o.second) ans[v[i]] = 1;
    }
    for (int i = n; i > 0; i--) {
      for (int j = i * 2; j <= n; j += i) ans[i] -= ans[j];
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  