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
  from sys import stdin
  
  from collections import deque
  def NC_Dij(lis,start):
  
      ret = [float("inf")] * len(lis)
      ret[start] = 0
      
      q = deque([start])
      plis = [i for i in range(len(lis))]
  
      while len(q) > 0:
          now = q.popleft()
  
          for nex in lis[now]:
  
              if ret[nex] > ret[now] + 1:
                  ret[nex] = ret[now] + 1
                  plis[nex] = now
                  q.append(nex)
      return ret,plis
  
  def search(k):
  
      enum = [len(lis[i]) for i in range(n)]
      able = [True] * n
  
      for d,v in dv:
          able[v] = False
          for nex in lis[v]:
              if able[nex] and enum[v] % k != 0:
                  enum[v] -= 1
              elif able[nex]:
                  enum[nex] -= 1
      
          if enum[v] % k != 0:
              return False
  
      return True
      
  mod = 998244353
  tt = int(stdin.readline())
  
  for loop in range(tt):
  
      n = int(stdin.readline())
  
      lis = [ [] for i in range(n) ]
  
      enum = [0] * n
      for i in range(n-1):
          u,v = map(int,stdin.readline().split())
          u -= 1
          v -= 1
          lis[u].append(v)
          lis[v].append(u)
          enum[u] += 1
          enum[v] += 1
  
      if n == 2:
          print (2,0)
          continue
  
      
      for i in range(n):
          if len(lis[i]) == 1:
              for nex in lis[i]:
                  enum[nex] -= 1
      
      mins = float("inf")
      for i in range(n):
          if enum[i] <= 1 and len(lis[i]) != 1:
              mins = len(lis[i]) - enum[i]
  
  
      dlis,tmp = NC_Dij(lis,0)
      dv = [(dlis[i],i) for i in range(n)]
      dv.sort()
      dv.reverse()
  
      ps = set()
      for i in range(1,int(mins**0.5+10)):
          if mins % i == 0:
              ps.add(i)
              ps.add(mins//i)
          if (mins+1) % i == 0:
              ps.add(i)
              ps.add((mins+1)//i)
  
      ans = [0] * (n+1)
      for p in ps:
          if p != 1 and (n-1) % p == 0 and search(p):
              ans[p] = 1
  
      for i in range(n+1):
          if ans[i] == 1:
              for j in range(2*i,n+1,i):
                  if ans[j] == 1:
                      ans[i] = 0
                      break
  
      ans[1] = ( pow(2,n-1,mod) - sum(ans) ) % mod
      print (*ans[1:])
      
      
  