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: import sys
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  MOD = 998244353
  
  t = int(input())
  while t > 0:
      t -= 1
      n = int(input())
      g = [[] for i in range(n)]
      for i in range(n - 1):
          x, y = map(int, input().split())
          g[x - 1] += [y - 1]
          g[y - 1] += [x - 1]
      f = [0] * n
      parent = [0] * n
      f[1] = pow(2, n - 1, MOD)
      order = [0]
      for v in order:
          for u in g[v]:
              if u != parent[v]:
                  parent[u] = v
                  order += [u]
  
  
      def dfs(k):
          size = [0] * n
          for v in reversed(order):
              if size[v] % k == 0:
                  if v != 0:
                      size[parent[v]] += 1
              elif v == 0 or (size[v] + 1) % k != 0:
                  return False
          return True
  
  
      for i in range(2, n):
          if (n - 1) % i == 0:
              f[i] = int(dfs(i))
      h = [0] * (n + 1)
      for i in range(n - 1, 0, -1):
          h[i] = f[i]
          for j in range(i * 2, n, i):
              h[i] -= h[j]
      print(*(x for x in h[1:n + 1]))
  