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;
  inline bool isdigit(char& ch) { return ch >= '0' && ch <= '9'; }
  template <class T>
  void read(T& x) {
    x = 0;
    long long f = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar())
      if (ch == '-') f = -1;
    for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
    x *= f;
  }
  template <class T>
  inline void write(T x) {
    if (x == 0) {
      putchar('0');
      return;
    }
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    int _stk[65], _top = 0;
    for (; x; x /= 10) _stk[++_top] = x % 10 + 48;
    for (; _top; _top--) putchar(_stk[_top]);
  }
  const int MAXN = 1e5 + 10;
  const int mod = 998244353;
  int f[MAXN], ans[MAXN], n, flag;
  vector<int> g[MAXN];
  void dfs(int u, int fa, int d) {
    if (!flag) return;
    for (auto v : g[u]) {
      if (v == fa) continue;
      dfs(v, u, d);
    }
    if (f[u] % d == 0)
      f[fa]++;
    else {
      if (fa) f[u]++;
      if (f[u] % d != 0) flag = 0;
    }
    if (!flag) return;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T;
    cin >> T;
    while (T--) {
      cin >> n;
      for (int i = (int)(1); i < (int)(n); i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
      }
      ans[1] = 1;
      for (int i = (int)(1); i < (int)(n); i++)
        ans[1] = (1ll * ans[1] + ans[1]) % mod;
      for (int d = (int)(2); d < (int)(n + 1); d++) {
        if ((n - 1) % d == 0) {
          flag = 1;
          dfs(1, 0, d);
          ans[d] = flag;
          for (int i = (int)(1); i < (int)(n + 1); i++) f[i] = 0;
        }
      }
      for (int i = n; i >= 1; --i) {
        for (int j = i + i; j <= n; j += i) ans[i] = (ans[i] - ans[j]) % mod;
      }
      for (int i = (int)(1); i < (int)(n + 1); i++) {
        ans[i] = (ans[i] % mod + mod) % mod;
        cout << ans[i] << ' ';
      }
      cout << '\n';
      for (int i = (int)(1); i < (int)(n + 1); i++) g[i].clear(), ans[i] = 0;
    }
    return 0;
  }
  