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 int N = 300000, mod = 998244353;
  int n;
  struct side {
    int y, next;
  } e[N * 2 + 9];
  int lin[N + 9], cs;
  void Ins(int x, int y) {
    e[++cs].y = y;
    e[cs].next = lin[x];
    lin[x] = cs;
  }
  void Ins2(int x, int y) {
    Ins(x, y);
    Ins(y, x);
  }
  void into() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) lin[i] = 0;
    cs = 0;
    for (int i = 1; i < n; ++i) {
      int x, y;
      scanf("%d%d", &x, &y);
      Ins2(x, y);
    }
  }
  int cnt[N + 9];
  void Dfs_cnt(int k, int fa, int p) {
    cnt[k] = 0;
    for (int i = lin[k]; i; i = e[i].next) {
      int y = e[i].y;
      if (y == fa) continue;
      Dfs_cnt(y, k, p);
      cnt[y] % p ? ++cnt[y] : ++cnt[k];
    }
  }
  int ans[N + 9];
  void Get_ans() {
    for (int i = 1; i <= n; ++i) ans[i] = 0;
    ans[1] = 1;
    for (int i = 1; i < n; ++i) ans[1] = 1LL * ans[1] * 2 % mod;
    for (int i = 2; i < n; ++i) {
      if ((n - 1) % i) continue;
      Dfs_cnt(1, 0, i);
      ans[i] = 1;
      for (int j = 1; j <= n; ++j)
        if (cnt[j] % i) ans[i] = 0;
    }
    for (int i = n; i >= 1; --i)
      for (int j = i << 1; j <= n; j += i) ans[i] = (ans[i] + mod - ans[j]) % mod;
  }
  void work() { Get_ans(); }
  void outo() {
    for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
    puts("");
  }
  int main() {
    int T = 1;
    scanf("%d", &T);
    for (; T--;) {
      into();
      work();
      outo();
    }
    return 0;
  }
  