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;
  typedef const int& ci;
  vector<int> sn[100000 + 1];
  int deg[100000 + 1], que[100000 + 1], par[100000 + 1], ans[100000],
      cnt[100000 + 1];
  int check(ci n, ci k) {
    memset(cnt + 1, 0, n << 2);
    for (int i = 1; i < n; ++i) {
      int u = que[i];
      if (cnt[u] % k) {
        if ((++cnt[u]) % k) return 0;
      } else
        ++cnt[par[u]];
    }
    return 1;
  }
  __inline__ __attribute__((always_inline)) void test() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) sn[i].clear();
    memset(deg + 1, 0, n << 2);
    for (int i = n, u, v; --i;) {
      scanf("%d%d", &u, &v);
      sn[u].push_back(v), sn[v].push_back(u), ++deg[u], ++deg[v];
    }
    int tail = 0;
    for (int i = 1; i <= n; ++i)
      if (deg[i] == 1) que[++tail] = i;
    for (int head = 1; head < n; ++head) {
      int u = que[head];
      for (auto v : sn[u])
        if (deg[v] > 1) {
          par[u] = v;
          if (--deg[v] == 1) que[++tail] = v;
        }
    }
    int m = n - 1;
    ans[1] = 1;
    for (int i = m; i; --i, ans[1] = (ans[1] << 1) % 998244353)
      ;
    for (int i = m; i > 1; --i)
      if (!(m % i)) {
        ans[i] = check(n, i);
        for (int j = i << 1; j <= m; j += i) ans[i] -= ans[j];
        ans[1] -= ans[i];
      } else
        ans[i] = 0;
    for (int i = 1; i < n; ++i) printf("%d ", ans[i]);
    puts("0");
  }
  int main() {
    int t;
    for (scanf("%d", &t); t; test(), --t)
      ;
    return 0;
  }
  