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;
  int n, a, b, g;
  int res[1000007];
  struct wierzch {
    list<int> l;
  };
  wierzch tab[1000007];
  int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int testuj(int nr, int o, int k) {
    int d = 0;
    for (auto el : tab[nr].l)
      if (el != o) {
        int x = testuj(el, nr, k);
        if (x == -1) return -1;
        d += x;
      }
    if (d % k == 0) {
      g = gcd(d, g);
      return 1;
    }
    if ((d + 1) % k == 0) {
      g = gcd(d + 1, g);
      return 0;
    }
    return -1;
  }
  int solve() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) tab[i].l.clear();
    for (int i = 0; i < n - 1; i++) {
      scanf("%d %d", &a, &b);
      tab[a].l.push_back(b);
      tab[b].l.push_back(a);
    }
    int r = 1;
    for (int i = 0; i < n - 1; i++) r = r * 2 % 998244353;
    for (int k = 2; k <= n; k++) {
      g = 0;
      if ((n - 1) % k == 0 && testuj(1, 1, k) == 1 && g == k)
        res[k] = 1;
      else
        res[k] = 0;
      r -= res[k];
    }
    res[1] = (r + 998244353) % 998244353;
    for (int i = 1; i <= n; i++) printf("%d ", res[i]);
    printf("\n");
    return 0;
  }
  int main() {
    int _tests;
    scanf("%d", &_tests);
    for (int _test = 1; _test <= _tests; _test++) solve();
    return 0;
  }
  