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 maxn = 300000 + 50;
  const long long mm = 998244353;
  int n, m, ans, T;
  long long f[maxn], w[maxn];
  vector<int> tr[maxn];
  long long dfs1(int x, int fa) {
    long long tot = 1;
    w[x] = 1;
    for (int i = 0; i < tr[x].size(); ++i)
      if (tr[x][i] != fa)
        tot = tot * dfs1(tr[x][i], x) * 2 % mm, w[x] = (w[x] + w[tr[x][i]]);
    return tot;
  }
  int dfs(int x, int fa, int z, int y) {
    int tot = 0;
    for (int i = 0; i < tr[x].size(); ++i)
      if (tr[x][i] != fa)
        if (w[tr[x][i]] % z == 0) {
          if (!dfs(tr[x][i], x, z, 1)) return 0;
        } else if ((w[tr[x][i]] - 1) % z == 0) {
          ++tot;
          if (!dfs(tr[x][i], x, z, 0)) return 0;
        } else
          return 0;
    if ((tot + y) % z == 0)
      ;
    else
      return 0;
    return 1;
  }
  int main() {
    scanf("%d", &T);
    for (int oo = 1; oo <= T; ++oo) {
      scanf("%d", &n);
      for (int i = 1; i <= n; ++i)
        while (!tr[i].empty()) tr[i].pop_back();
      for (int i = 1; i < n; ++i) {
        int q1, q2;
        scanf("%d%d", &q1, &q2);
        tr[q1].push_back(q2);
        tr[q2].push_back(q1);
      }
      for (int i = 1; i <= n; ++i) f[i] = 0;
      f[1] = dfs1(1, 1);
      for (int i = n - 1; i > 1; --i)
        if ((n - 1) % i == 0 && !f[i]) {
          f[i] += dfs(1, 1, i, 0);
          if (f[i])
            for (int j = 1; j * j <= i; ++j)
              if (i % j == 0) {
                f[j] -= f[i];
                if (j * j != i && j != 1) f[i / j] -= f[i];
              }
        }
      for (int i = 1; i <= n; ++i)
        if (f[i] >= 0)
          printf("%lld ", (f[i] + mm) % mm);
        else
          printf("0 ");
      printf("\n");
    }
    return 0;
  }
  