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>
  #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
  using namespace std;
  const long long N = 1e5 + 100;
  vector<long long> p[N];
  vector<long long> g[N];
  long long long long ans[N];
  long long long long mod = 998244353;
  long long long long dp[N];
  long long long long vis[N];
  long long long long fac[N];
  long long long long fm[N];
  long long long long sz[N];
  long long long long power(long long long long n, long long long long k) {
    if (k == 0) return 1;
    if (k % 2 == 1) {
      long long long long x = power(n, k / 2);
      return x * x % mod * n % mod;
    }
    long long long long x = power(n, k / 2);
    return x * x % mod;
  }
  long long long long ent(long long long long k, long long long long n) {
    if (k < 0 || k > n) return 0;
    if (k == 0 || k == n) return 1;
    return fac[n] * fm[k] % mod * fm[n - k] % mod;
  }
  void dfs(long long long long v, long long long long p, long long long long x) {
    sz[v] = 1;
    vis[v] = 0;
    long long long long z = g[v].size();
    long long long long p1 = 0, p2 = 0;
    for (auto u : g[v]) {
      if (u == p) continue;
      dfs(u, v, x);
      if (vis[u])
        z--, p1++;
      else
        p2++;
      sz[v] += sz[u];
    }
    if (z % x == 0) vis[v] = 1;
    if (z % x > 1) {
      dp[v] = 0;
      return;
    }
    dp[v] = 1;
    long long long long jj = sz[v];
    for (auto u : g[v]) {
      if (u == p) continue;
      dp[v] *= dp[u];
      dp[v] %= mod;
    }
  }
  long long Main() {
    long long long long n;
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      g[i].clear();
      ans[i] = 0;
    }
    for (long long i = 0; i < n - 1; i++) {
      long long long long u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    ans[1] = 1;
    for (long long i = 1; i < n; i++) ans[1] = ans[1] * 2 % mod;
    long long long long id = 1;
    for (long long i = 1; i <= n; i++) {
      if (g[i].size() == 1) continue;
      long long long long p1 = 0;
      for (auto u : g[i]) {
        if (g[u].size() > 1) p1++;
      }
      if (p1 <= 1) {
        id = g[i].size();
      }
    }
    for (auto u : p[id]) {
      dfs(1, 1, u);
      ans[u] = dp[1];
      if (vis[1] == 0) ans[u] = 0;
    }
    for (auto u : p[id - 1]) {
      dfs(1, 1, u);
      ans[u] = dp[1];
      if (vis[1] == 0) ans[u] = 0;
    }
    for (long long i = n; i; i--) {
      for (long long j = i + i; j <= n; j += i) {
        ans[i] -= ans[j];
        ans[i] += mod;
        ans[i] %= mod;
      }
    }
    for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << endl;
    return 0;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    fac[0] = 1;
    fm[0] = 1;
    for (long long i = 1; i < N; i++) {
      fac[i] = fac[i - 1] * i % mod;
      fm[i] = power(fac[i], mod - 2);
    }
    for (long long i = 2; i < N; i++) {
      for (long long j = i; j < N; j += i) p[j].push_back(i);
    }
    long long long long t = 1;
    cin >> t;
    while (t--) Main();
  }
  