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;
  long long M = 998244353;
  long long fpow(long long x, long long y) {
    if (y == 0) return 1;
    if (y == 1) return x;
    long long p = fpow(x, y / 2);
    p *= p;
    p %= M;
    if (y & 1) p *= x;
    return p % M;
  }
  vector<int> v[100005];
  long long ans[100005], po[100005];
  bool ok;
  int dfs(int ver, int k, int par) {
    int sum = 0;
    for (int i = 0; i < v[ver].size(); i++) {
      if (v[ver][i] == par) continue;
      sum += dfs(v[ver][i], k, ver);
    }
    if (sum % k == 0) {
      return 1;
    }
    if ((sum + 1) % k != 0) ok = 0;
    return 0;
  }
  int main() {
    po[0] = 1;
    for (int i = 1; i < 100005; i++) {
      po[i] = 2 * po[i - 1];
      po[i] %= M;
    }
    int t, x, y, n;
    cin >> t;
    while (t--) {
      cin >> n;
      for (int i = 0; i < n - 1; i++) {
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
      }
      ans[1] = po[n - 1];
      for (int k = 2; k <= n; k++) {
        ok = 1;
        ans[k] = 0;
        if ((n - 1) % k != 0) continue;
        dfs(1, k, -1);
        ans[k] = ok;
      }
      for (int k = n; k >= 1; k--) {
        for (int i = k * 2; i <= n; i += k) {
          ans[k] = (ans[k] - ans[i] + M) % M;
        }
      }
      for (int i = 1; i <= n; i++) {
        cout << ans[i] << ' ';
        v[i].clear();
      }
      cout << endl;
    }
    return 0;
  }
  