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;
  void dbg() { std::cout << "  #\n"; }
  template <typename T, typename... Args>
  void dbg(T a, Args... args) {
    std::cout << a << ' ';
    dbg(args...);
  }
  const long long maxn = 2e5 + 10;
  const long long MAX = 1000;
  const long long inf = 0x3f3f3f3f;
  const long long mod = 998244353;
  long long n, m;
  long long a[maxn];
  long long f[maxn];
  long long ans[maxn];
  vector<long long> A[maxn];
  long long getint() {
    char ch;
    while (!isdigit(ch = getchar()))
      ;
    long long x = ch - 48;
    while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
    return x;
  }
  bool isok;
  void dfs(long long now, long long fa, long long k) {
    for (auto to : A[now])
      if (to != fa) {
        dfs(to, now, k);
      }
    if (a[now] % k == 0) {
      a[fa]++;
    } else {
      if (fa != 0) {
        a[now]++;
        if (a[now] % k != 0) {
          isok = false;
        }
      }
    }
  }
  void solve() {
    n = getint();
    for (long long i = (1); i <= (n); ++i) ans[i] = 0, A[i].clear();
    for (long long i = (1); i <= (n - 1); ++i) {
      long long x, y;
      x = getint();
      y = getint();
      A[x].push_back(y);
      A[y].push_back(x);
    }
    ans[1] = 1;
    for (long long i = (1); i <= (n - 1); ++i) ans[1] = (ans[1] * 2ll % mod);
    for (long long i = (2); i <= (n); ++i)
      if ((n - 1) % i == 0) {
        for (long long j = (1); j <= (n); ++j) a[j] = 0;
        isok = 1;
        dfs(1, 0, i);
        if (isok) ans[i]++;
      }
    for (long long i = (n); i >= (1); --i) {
      long long j = i + i;
      while (j <= n) {
        ans[i] -= ans[j];
        ans[i] = (ans[i] + mod) % mod;
        j = j + i;
      }
    }
    for (long long i = (1); i <= (n); ++i) printf("%d ", ans[i] % mod);
    printf("\n");
  }
  signed main() {
    long long TestCase = 1;
    cin >> TestCase;
    while (TestCase--) {
      solve();
    }
    char EndFile = getchar();
    EndFile = getchar();
    return 0;
  }
  