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, k, dp[100010], fl, ans[100010];
  vector<int> g[100010];
  const int mod = 998244353;
  void dfs(int x, int f) {
    int s = 0;
    for (int i = 0, v; i < g[x].size(); i++) {
      v = g[x][i];
      if (v == f) continue;
      dfs(v, x);
      s += dp[v];
    }
    dp[x] = 1;
    if (f && s % k) s++, dp[x] = 0;
    if (s % k) fl = 1;
  }
  const int N = 1e5;
  int mu[N + 10], f[N + 10], pri[N + 10], cn;
  void sieve() {
    mu[1] = 1;
    for (int i = 2; i <= N; i++) {
      if (!f[i]) pri[++cn] = i, mu[i] = -1;
      for (int j = 1; j <= cn && i * pri[j] <= N; j++) {
        f[i * pri[j]] = 1;
        if (!(i % pri[j])) break;
        mu[i * pri[j]] = -mu[i];
      }
    }
  }
  void sol() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) g[i].clear(), ans[i] = dp[i] = f[i] = 0;
    for (int i = 1, u, v; i < n; i++) {
      scanf("%d%d", &u, &v);
      g[u].push_back(v);
      g[v].push_back(u);
    }
    f[1] = 1;
    for (int i = 1; i < n; i++) (f[1] *= 2) %= mod;
    for (k = 2; k < n; k++)
      if (!((n - 1) % k)) {
        fl = 0;
        dfs(1, 0);
        if (!fl) f[k] = 1;
      }
    for (int i = 1; i <= n; i++)
      for (int j = 1; i * j <= n; j++) (ans[i] += f[i * j] * mu[j]) %= mod;
    for (int i = 1; i <= n; i++) printf("%d ", (ans[i] + mod) % mod);
    printf("\n");
  }
  int main() {
    sieve();
    int T;
    scanf("%d", &T);
    while (T--) sol();
    return 0;
  }
  