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 target("avx2")
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  using ll = long long;
  using ld = long double;
  const ll MOD = 998244353;
  const ll root = 3;
  ll binpow(ll a, ll b) {
    ll res = 1;
    while (b) {
      if (b & 1) res = (res * a) % MOD;
      a = (a * a) % MOD;
      b >>= 1;
    }
    return res;
  }
  ll modInv(ll a) { return binpow(a, MOD - 2); }
  const int INF = 0x3f3f3f3f;
  const int NINF = 0xc0c0c0c0;
  const ll INFLL = 0x3f3f3f3f3f3f3f3f;
  const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
  const int mxN = 100001;
  int n;
  vector<int> adj[mxN];
  int ans[mxN];
  ll calc(int u, int p = -1) {
    ll curr = 1;
    for (int v : adj[u]) {
      if (v ^ p) {
        curr = (curr * 2LL * calc(v, u)) % MOD;
      }
    }
    return curr;
  }
  bool die = false;
  int go(int u, int k, int p = -1) {
    int curr = 0;
    for (int v : adj[u]) {
      if (v ^ p) {
        curr += go(v, k, u);
      }
    }
    if (curr % k == 0)
      return 1;
    else if ((curr + 1) % k == 0)
      return 0;
    die = 1;
    return 0;
  }
  int check(int k) {
    if (k == 1) {
      return calc(0);
    }
    for (int i = (0); i < (n); ++i) {
      if (adj[i].size() > 1) {
        die = false;
        bool ok = go(i, k);
        if (die == 0 && ok == 1) return 1;
        return 0;
      }
    }
  }
  void solve() {
    cin >> n;
    for (int i = (0); i < (n + 1); ++i) {
      ans[i] = 0;
      adj[i].clear();
    }
    for (int i = (0); i < (n - 1); ++i) {
      int a, b;
      cin >> a >> b;
      a--;
      b--;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    if (n == 2) {
      cout << "2 0\n";
      return;
    }
    for (int d = 1; d * d <= n - 1; d++) {
      if ((n - 1) % d == 0) {
        ans[d] = check(d);
        ans[(n - 1) / d] = check((n - 1) / d);
      }
    }
    for (int i = (n)-1; i >= (1); --i) {
      for (int j = i + i; j < n; j += i) {
        ans[i] = (ans[i] - ans[j] + MOD) % MOD;
      }
    }
    for (int i = (1); i < (n + 1); ++i) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
  int main() {
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    cin.tie(0)->sync_with_stdio(0);
    int T = 1;
    cin >> T;
    for (int tc = 1; tc <= T; tc++) {
      solve();
    }
    return 0;
  }
  