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>
  const int MAXN = 1e5, MOD = 998244353;
  int n, ecnt, head[MAXN + 5], f[MAXN + 5];
  struct Edge {
    int to, nxt;
  } graph[MAXN * 2 + 5];
  inline void subeq(int& a, const int b) { (a -= b) < 0 && (a += MOD); }
  inline int mul(const int a, const int b) { return int(1ll * a * b % MOD); }
  inline int mpow(int a, int b) {
    int ret = 1;
    for (; b; a = mul(a, a), b >>= 1) ret = mul(ret, b & 1 ? a : 1);
    return ret;
  }
  inline void link(const int u, const int v) {
    graph[++ecnt] = {v, head[u]}, head[u] = ecnt;
    graph[++ecnt] = {u, head[v]}, head[v] = ecnt;
  }
  inline bool check(const int u, const int fa, const int k) {
    static int val[MAXN + 5];
    val[u] = 0;
    for (int i = head[u], v; i; i = graph[i].nxt) {
      if ((v = graph[i].to) != fa) {
        if (!check(v, u, k)) return false;
      }
    }
    if (val[u] % k) {
      if (!fa || (val[u] + 1) % k) return false;
    } else if (fa)
      ++val[fa];
    return true;
  }
  inline void allClear() {
    ecnt = 0;
    for (int i = 1, repi = n; i <= repi; ++i) head[i] = 0;
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &n), allClear();
      for (int i = 2, repi = n; i <= repi; ++i) {
        int u, v;
        scanf("%d %d", &u, &v);
        link(u, v);
      }
      f[1] = mpow(2, n - 1);
      for (int i = 2, repi = n - 1; i <= repi; ++i)
        f[i] = !((n - 1) % i) && check(1, 0, i);
      f[n] = 0;
      for (int i = n, peri = 1; i >= peri; --i)
        for (int j = 2, repj = n / i; j <= repj; ++j) subeq(f[i], f[i * j]);
      for (int i = 1, repi = n; i <= repi; ++i)
        printf("%d%c", f[i], i < n ? ' ' : '\n');
    }
    return 0;
  }
  