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;
  using ll = long long;
  using ld = long double;
  using db = double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using pd = pair<db, db>;
  using vi = vector<int>;
  using vb = vector<bool>;
  using vl = vector<ll>;
  using vd = vector<db>;
  using vs = vector<str>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  using vpd = vector<pd>;
  ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
  ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T>
  void remDup(vector<T>& v) {
    sort(begin(v), end(v));
    v.erase(unique(begin(v), end(v)), v.end());
  }
  const int MOD = 998244353;
  const int MX = 1e5 + 10;
  const ll INF = 1e18;
  struct mint {
    int v;
    explicit operator int() const { return v; }
    mint() { v = 0; }
    mint(ll _v) {
      v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
      if (v < 0) v += MOD;
    }
    friend bool operator==(const mint& a, const mint& b) { return a.v == b.v; }
    friend bool operator!=(const mint& a, const mint& b) { return !(a == b); }
    friend bool operator<(const mint& a, const mint& b) { return a.v < b.v; }
    mint& operator+=(const mint& m) {
      if ((v += m.v) >= MOD) v -= MOD;
      return *this;
    }
    mint& operator-=(const mint& m) {
      if ((v -= m.v) < 0) v += MOD;
      return *this;
    }
    mint& operator*=(const mint& m) {
      v = int((ll)v * m.v % MOD);
      return *this;
    }
    mint& operator/=(const mint& m) { return (*this) *= inv(m); }
    friend mint pow(mint a, long long p) {
      mint ans = 1;
      for (; p; p /= 2, a *= a)
        if (p & 1) ans *= a;
      return ans;
    }
    friend mint inv(const mint& a) { return pow(a, MOD - 2); }
    mint operator-() const { return mint(-v); }
    mint& operator++() { return *this += 1; }
    mint& operator--() { return *this -= 1; }
    friend mint operator+(mint a, const mint& b) { return a += b; }
    friend mint operator-(mint a, const mint& b) { return a -= b; }
    friend mint operator*(mint a, const mint& b) { return a *= b; }
    friend mint operator/(mint a, const mint& b) { return a /= b; }
  };
  int T, N;
  vi adj[MX];
  mint DP[MX];
  int cnt[MX], DIV;
  bool bad;
  void DFS(int X, int P) {
    if (bad) return;
    for (auto& Y : adj[X])
      if (Y != P) DFS(Y, X);
    if (cnt[X] % DIV != 0) {
      if (P != 0) cnt[X]++;
      if (cnt[X] % DIV != 0) {
        bad = true;
        return;
      }
    } else
      cnt[P]++;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> T;
    while (T--) {
      cin >> N;
      for (int i = (1); i < (N + 1); i++) adj[i].clear(), DP[i] = 0;
      for (int i = (0); i < (N - 1); i++) {
        int A, B;
        cin >> A >> B;
        adj[A].push_back(B), adj[B].push_back(A);
      }
      DP[1] = pow(mint(2), N - 1);
      for (int i = (2); i < (N + 1); i++)
        if ((N - 1) % i == 0) {
          bad = false;
          DIV = i;
          for (int j = (1); j < (N + 1); j++) cnt[j] = 0;
          DFS(1, 0);
          if (!bad) DP[i] = 1;
        }
      for (int i = (N + 1) - 1; i >= (1); i--)
        for (int j = 2 * i; j <= N; j += i) DP[i] -= DP[j];
      for (int i = (1); i < (N + 1); i++) cout << int(DP[i]) << " ";
      cout << "\n";
    }
  }
  