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;
  vector<long long> ch[100005];
  long long num[100005], B = 998244353;
  long long dfs(long long now, long long from, long long mod) {
    long long a, b, c;
    c = 0;
    for (auto a : ch[now]) {
      if (a == now || a == from) continue;
      b = dfs(a, now, mod);
      if (b == -1) return -1;
      c += b;
    }
    if ((ch[now].size() - c) % mod == 0) return 1;
    if (ch[now].size() - c > 0 && (ch[now].size() - c - 1) % mod == 0 &&
        from != 0)
      return 0;
    return -1;
  }
  void solve() {
    long long i, j, k, a, b, c, n;
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) num[i] = 0;
    for (i = 1; i <= n; i++) {
      ch[i].clear();
    }
    for (i = 1; i < n; i++) {
      scanf("%lld%lld", &a, &b);
      ch[a].push_back(b);
      ch[b].push_back(a);
    }
    for (i = 2; i <= n; i++) {
      if ((2 * n - 2) % i) continue;
      if (dfs(1, 0, i) >= 0) num[i] = 1;
    }
    num[1] = 1;
    for (i = 1; i < n; i++) num[1] = (num[1] * 2) % B;
    c = 0;
    for (i = n; i > 1; i--) {
      for (j = i * 2; j <= n; j += i) {
        if (num[j] == 1) num[i] = 0;
      }
      c += num[i];
    }
    num[1] -= c;
    while (num[1] < 0) num[1] += B;
    for (i = 1; i <= n; i++) printf("%lld ", num[i]);
    printf("\n");
  }
  int main() {
    int T = 1;
    scanf("%d", &T);
    while (T--) {
      solve();
    }
  }
  