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>
  template <typename T>
  bool chkmin(T &a, T b) {
    return (b < a) ? a = b, 1 : 0;
  }
  template <typename T>
  bool chkmax(T &a, T b) {
    return (b > a) ? a = b, 1 : 0;
  }
  long long ksm(long long a, long long b) {
    if (b == 0) return 1;
    long long ns = ksm(a, b >> 1);
    ns = ns * ns % 998244353;
    if (b & 1) ns = ns * a % 998244353;
    return ns;
  }
  using namespace std;
  const int maxn = 100050;
  int n;
  vector<int> g[maxn];
  int cn[maxn];
  int ans[maxn];
  void dfs(int u, int p, int k) {
    int deg = 0;
    for (int v : g[u]) {
      if (v != p) {
        deg++;
        dfs(v, u, k);
        if (cn[v] % k != 0) {
          cn[v]++;
          deg--;
        }
      }
    }
    cn[u] = deg;
  }
  int check(int k) {
    for (int i = 1; i <= n; i++) cn[i] = 0;
    dfs(1, 0, k);
    for (int i = 1; i <= n; i++) {
      if (cn[i] % k != 0) {
        return 0;
      }
    }
    return 1;
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; i++) g[i].clear();
      for (int i = 1; i < n; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        g[u].push_back(v);
        g[v].push_back(u);
      }
      ans[1] = ksm(2, n - 1);
      for (int k = 2; k <= n; k++) {
        if ((n - 1) % k != 0) {
          ans[k] = 0;
          continue;
        }
        ans[k] = check(k);
      }
      for (int i = n; i >= 1; i--) {
        for (int j = i + i; j <= n; j += i) {
          ans[i] -= ans[j];
        }
      }
      for (int i = 1; i <= n; i++) {
        printf("%d ", ans[i]);
      }
      printf("\n");
    }
  }
  