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>
  const int mod = 998244353;
  using std::vector;
  int ans[100000 + 5];
  vector<int> e[100000 + 5];
  int qpow(int bsc, int y) {
    int ret = 1;
    while (y) {
      if (y & 1) ret = (long long)ret * bsc % mod;
      bsc = (long long)bsc * bsc % mod;
      y >>= 1;
    }
    return ret;
  }
  bool flag;
  int d;
  int dfs(int u, int f) {
    int cnt = 0;
    for (int v : e[u]) {
      if (v == f) continue;
      int sum = dfs(v, u);
      if (!flag) return 0;
      if (sum % d && (sum + 1) % d) {
        flag = false;
        return 0;
      }
      cnt += sum + 1;
    }
    return cnt;
  }
  bool check(int _d) {
    flag = true;
    d = _d;
    dfs(1, 0);
    return flag;
  }
  void solve() {
    int n;
    scanf("%d", &n);
    for (int i = 2; i <= n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      e[u].push_back(v);
      e[v].push_back(u);
    }
    for (int i = 1; i <= n; i++) ans[i] = 0;
    int cnto = 0;
    vector<int> fs;
    for (int i = 2; i * i <= n - 1; i++)
      if ((n - 1) % i == 0) {
        fs.push_back(i);
        if (i * i != n - 1) fs.push_back((n - 1) / i);
      }
    if (n - 1 != 1) fs.push_back(n - 1);
    for (int d : fs) {
      if (check(d)) {
        cnto++;
        ans[d] = 1;
      }
    }
    for (int x : fs) {
      if (!ans[x]) continue;
      for (int y : fs) {
        if (!ans[y]) continue;
        if (x != y && x % y == 0) {
          ans[y] = 0;
          cnto--;
        }
      }
    }
    ans[1] = (qpow(2, n - 1) + mod - cnto) % mod;
    for (int i = 1; i <= n; i++) printf("%d%c", ans[i], " \n"[i == n]);
    for (int i = 1; i <= n; i++) {
      e[i].clear();
    }
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) solve();
  }
  