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 T, n, ans[200005], val[200005];
  int cnt, pri[200005], vis[200005], mu[200005];
  vector<int> g[200005];
  void init(int lim) {
    mu[1] = 1;
    for (int i = 2; i <= lim; ++i) {
      if (!vis[i]) pri[++cnt] = i, mu[i] = -1;
      for (int j = 1; j <= cnt && pri[j] * i <= lim; ++j) {
        vis[pri[j] * i] = 1;
        if (i % pri[j] == 0) {
          mu[i * pri[j]] = 0;
          break;
        }
        mu[pri[j] * i] = -mu[i];
      }
    }
  }
  int ksm(int x, int y) {
    int num = 1;
    while (y) {
      if (y & 1) num = 1LL * num * x % 998244353;
      x = 1LL * x * x % 998244353, y >>= 1;
    }
    return num;
  }
  bool dfs(int x, int fa, int k) {
    val[x] = 0;
    int flag = 1, num = 0;
    for (int to : g[x])
      if (to != fa) {
        if (!dfs(to, x, k)) flag = 0;
        if (!val[to]) num++;
      }
    if (num % k == 0)
      val[x] = 0;
    else if ((num + 1) % k == 0)
      val[x] = 1;
    else
      flag = 0;
    return flag;
  }
  int main() {
    scanf("%d", &T);
    init(1e5);
    while (T--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; ++i) {
        ans[i] = 0;
        g[i].clear();
      }
      for (int i = 1; i < n; ++i) {
        int x, y;
        scanf("%d%d", &x, &y);
        g[x].push_back(y), g[y].push_back(x);
      }
      for (int i = 1; i * i <= (n - 1); ++i)
        if ((n - 1) % i == 0) {
          ans[i] = dfs(1, 0, i);
          if (i * i != n - 1) ans[(n - 1) / i] = dfs(1, 0, (n - 1) / i);
        }
      ans[1] = ksm(2, n - 1);
      for (int i = 1; i <= n; ++i) {
        for (int j = 2; j * i <= n; ++j)
          ans[i] = (ans[i] + mu[j] * ans[i * j] + 998244353) % 998244353;
        printf("%d ", ans[i]);
      }
      printf("\n");
    }
    return 0;
  }
  