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;
  int read() {
    char c = getchar();
    int s = 0;
    int x = 1;
    while (c < '0' || c > '9') {
      if (c == '-') x = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      s = s * 10 + c - '0';
      c = getchar();
    }
    return s * x;
  }
  const int mod = 998244353;
  const int N = 1e5 + 5;
  int T, n, x, y, f[N], ans[N];
  vector<int> v[N];
  int ksm(int x, int k) {
    int base = 1;
    while (k) {
      if (k & 1) base = 1ll * base * x % mod;
      k >>= 1;
      x = 1ll * x * x % mod;
    }
    return base;
  }
  int gcd(int x, int y) {
    if (!x) return y;
    return gcd(y % x, x);
  }
  void dfs(int u, int fa, int p) {
    int siz = 0;
    for (int i = 0; i < v[u].size(); i++) {
      if (v[u][i] == fa) continue;
      dfs(v[u][i], u, p);
      siz++;
    }
    if (!siz) {
      f[fa]++;
    } else {
      if (fa) {
        if (f[u] % p == 0)
          f[fa]++;
        else
          f[u]++;
      }
    }
  }
  int main() {
    T = read();
    while (T--) {
      n = read();
      for (int i = 1; i < n; i++) {
        x = read();
        y = read();
        v[x].push_back(y);
        v[y].push_back(x);
      }
      int tmp = ksm(2, n - 1);
      int pre = 0;
      for (int i = 2; i < n; i++) {
        if ((n - 1) % i == 0) {
          memset(f, 0, (n + 1) * 4);
          dfs(1, 0, i);
          int mi = 0;
          bool flag = 0;
          for (int k = 1; k <= n; k++) {
            mi = gcd(f[k], mi);
            if (f[k] % i) flag = 1;
          }
          if (!flag && mi == i) ans[i] = 1, pre++;
        }
      }
      cout << (tmp - pre + mod) % mod << " ";
      for (int i = 2; i <= n; i++) {
        cout << ans[i] << " ";
      }
      memset(ans, 0, (n + 1) * 4);
      for (int i = 1; i <= n; i++) v[i].clear();
      puts("");
    }
    return 0;
  }
  