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: from collections import deque
  import sys
  input = sys.stdin.readline
  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
  for loop in range(int(input())):
      n = int(input());lis = [ [] for i in range(n) ];enum = [0] * n
      for i in range(n-1):u,v = map(int,input().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();ans = [0] * (n+1)
      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)
      for p in ps:
          if p != 1 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:])