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;
  const int N = 1e6 + 10;
  const long long mod = 998244353;
  inline long long gcd(long long a, long long b) {
    return !b ? a : gcd(b, a % b);
  }
  inline long long q_pow(long long a, long long x) {
    long long ans = 1, tmp = a;
    while (x) {
      if (x & 1) (ans *= tmp) %= mod;
      (tmp *= tmp) %= mod;
      x >>= 1;
    }
    return ans;
  }
  template <typename T>
  inline void re(T &N) {
    int f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9')
      if (c == '-') f = -1;
    N = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') N = N * 10 + c - '0';
    N *= f;
  }
  template <class T, class... T_>
  inline void re(T &x, T_ &...y) {
    re(x), re(y...);
  }
  int m, n, t = 1, st, en;
  struct edge {
    int to, nxt;
  } e[N << 1];
  int head[N];
  inline void add(int x, int y) {
    e[++t] = {y, head[x]};
    head[x] = t;
    e[++t] = {x, head[y]};
    head[y] = t;
  }
  int a[N], ok = 1;
  void dfs(int x, int p, int k) {
    a[x] = 0;
    for (int i = head[x]; i; i = e[i].nxt) {
      int to = e[i].to;
      if (to == p) continue;
      dfs(to, x, k);
      if (a[to] % k)
        a[to]++;
      else
        a[x]++;
      if (a[to] % k) ok = 0;
    }
  }
  int check(int k) {
    ok = 1;
    dfs(1, 0, k);
    if (a[1] % k) ok = 0;
    return ok;
  }
  int p[N], v[N], ma_p[N], mi_p[N];
  int phi[N], miu[N];
  void get_prime() {
    phi[1] = miu[1] = 1;
    ma_p[1] = mi_p[1] = 1;
    for (int i = 2; i <= N - 1; i++) {
      if (!v[i])
        v[i] = i, p[++p[0]] = i, mi_p[i] = ma_p[i] = i, phi[i] = i - 1,
        miu[i] = -1;
      for (int j = 1; j <= p[0]; j++) {
        if (p[j] > v[i] || p[j] > (N - 1) / i) break;
        int nxt = i * p[j];
        v[nxt] = p[j];
        ma_p[nxt] = ma_p[i];
        mi_p[nxt] = p[j];
        phi[nxt] = phi[i] * (i % p[j] ? p[j] - 1 : p[j]);
        miu[nxt] = miu[i] * (i % p[j] ? -1 : 0);
      }
    }
  }
  long long f[N], g[N];
  int main() {
    get_prime();
    int T;
    re(T);
    while (T--) {
      re(n);
      t = 1;
      for (int i = 1; i <= n; i++) head[i] = f[i] = g[i] = 0;
      for (int i = 1; i < n; i++) {
        int x, y;
        re(x, y);
        add(x, y);
      }
      if (n == 2) {
        puts("2 0");
        continue;
      }
      f[1] = q_pow(2, n - 1);
      f[n - 1] = check(n - 1);
      for (int i = 2; i * i <= n - 1; i++)
        if (!((n - 1) % i)) {
          f[i] = check(i);
          f[(n - 1) / i] = check((n - 1) / i);
        }
      for (int i = 1; i <= n; i++)
        for (int j = i; j <= n; j += i) g[i] += miu[j / i] * f[j];
      for (int i = 1; i <= n; i++) printf("%lld ", (g[i] % mod + mod) % mod);
      puts("");
    }
  }
  