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;
  template <typename T>
  inline void read(T &a) {
    T w = 1;
    a = 0;
    char ch = getchar();
    for (; ch < '0' || ch > '9'; ch = getchar())
      if (ch == '-') w = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
      a = (a << 3) + (a << 1) + (ch ^ 48);
    a *= w;
  }
  template <typename T>
  inline void ckmax(T &a, T b) {
    a = a > b ? a : b;
  }
  template <typename T>
  inline void ckmin(T &a, T b) {
    a = a < b ? a : b;
  }
  const int N = 1e5 + 10;
  const int Mod = 998244353;
  vector<int> to[N];
  int f[N], g[N];
  inline int Sub(int x, int y) { return (x - y) < 0 ? (x + Mod - y) : (x - y); }
  inline int Mul(int x, int y) {
    return (1ll * x * y) >= Mod ? (1ll * x * y % Mod) : (x * y);
  }
  int qpow(int a, int b) {
    int base = a, ans = 1;
    while (b) {
      if (b & 1) ans = Mul(ans, base);
      base = Mul(base, base);
      b >>= 1;
    }
    return ans;
  }
  int fl;
  int dfs(int u, int fa, int k) {
    int cnt = 0;
    for (auto v : to[u]) {
      if (v == fa) continue;
      cnt += dfs(v, u, k);
    }
    if (!(cnt % k)) return 1;
    if (!((cnt + 1) % k)) return 0;
    fl = 0;
    return 0;
  }
  int main() {
    int cases;
    read(cases);
    while (cases--) {
      int n;
      read(n);
      memset(f, 0, sizeof f);
      for (int i = 1, u, v; i < n; ++i) {
        read(u);
        read(v);
        to[u].push_back(v);
        to[v].push_back(u);
      }
      f[1] = qpow(2, n - 1);
      for (int i = 2; i < n; ++i) {
        if (!((n - 1) % i)) {
          fl = 1;
          f[i] = dfs(1, 0, i) && fl;
        }
      }
      for (int i = n - 1; i >= 1; --i) {
        for (int j = i + i; j < n; j += i) f[i] = Sub(f[i], f[j]);
      }
      for (int i = 1; i <= n; ++i) printf("%d ", f[i]);
      puts("");
      for (int i = 1; i <= n; ++i) to[i].clear();
    }
  }
  