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;
  template <typename T1, typename T2>
  void chkmin(T1 &x, T2 y) {
    if (x > y) x = y;
  }
  template <typename T1, typename T2>
  void chkmax(T1 &x, T2 y) {
    if (x < y) x = y;
  }
  namespace fastio {
  char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
  inline char getc() {
    return p1 == p2 &&
                   (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
               ? -1
               : *p1++;
  }
  inline void putc(char x) { (*p3++ = x); }
  template <typename T>
  void read(T &x) {
    x = 0;
    char c = getchar();
    T neg = 0;
    while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
    while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    if (neg) x = (~x) + 1;
  }
  template <typename T>
  void recursive_print(T x) {
    return (!x) ? void() : (recursive_print(x / 10), putc(x % 10 ^ 48), void());
  }
  template <typename T>
  void print(T x) {
    (!x) && (putc('0'), 0);
    (x < 0) && (putc('-'), x = ~x + 1);
    recursive_print(x);
  }
  template <typename T>
  void print(T x, char c) {
    (!x) && (putc('0'), 0);
    (x < 0) && (putc('-'), x = ~x + 1);
    recursive_print(x);
    putc(c);
  }
  void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
  }  // namespace fastio
  const int MOD = 998244353;
  const int MAXN = 1e5;
  int n, hd[MAXN + 5], to[MAXN * 2 + 5], nxt[MAXN * 2 + 5], ec = 0, deg[MAXN + 5],
                                                            pw2[MAXN + 5];
  void adde(int u, int v) {
    to[++ec] = v;
    nxt[ec] = hd[u];
    hd[u] = ec;
  }
  int res[MAXN + 5], vis[MAXN + 5], cdeg[MAXN + 5], lft[MAXN + 5];
  void solve() {
    scanf("%d", &n);
    ec = 0;
    for (int i = 1; i <= n; i++) hd[i] = deg[i] = res[i] = 0;
    int cnt = 0;
    for (int i = 1, u, v; i < n; i++)
      scanf("%d%d", &u, &v), deg[u]++, deg[v]++, adde(u, v), adde(v, u);
    for (int i = 2; i <= n - 1; i++) {
      if ((n - 1) % i) continue;
      queue<int> q;
      bool flg = 1;
      for (int j = 1; j <= n; j++) vis[j] = 0, cdeg[j] = lft[j] = deg[j];
      for (int j = 1; j <= n; j++)
        if (deg[j] == 1) q.push(j);
      while (!q.empty()) {
        int x = q.front();
        q.pop();
        vis[x] = 1;
        int y = 0;
        for (int e = hd[x]; e; e = nxt[e])
          if (!vis[to[e]]) y = to[e];
        if (!y) {
          if (cdeg[x] % i) {
            flg = 0;
            break;
          }
        } else {
          if (cdeg[x] % i == 0)
            cdeg[y]--;
          else if (cdeg[x] % i == 1)
            cdeg[x]--;
          else {
            flg = 0;
            break;
          }
          if ((--lft[y]) <= 1) q.push(y);
        }
      }
      res[i] = flg;
      cnt += flg;
    }
    res[1] = pw2[n - 1];
    for (int i = n; i; i--)
      for (int j = i << 1; j <= n; j += i) res[i] = (res[i] - res[j] + MOD) % MOD;
    for (int i = 1; i <= n; i++) printf("%d%c", res[i], " \n"[i == n]);
  }
  int main() {
    for (int i = (pw2[0] = 1); i <= MAXN; i++) pw2[i] = (pw2[i - 1] << 1) % MOD;
    int qu;
    scanf("%d", &qu);
    while (qu--) solve();
    return 0;
  }
  