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>
  #pragma comment(linker, "/STACK:150777216")
  using namespace std;
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  mt19937 rnd(123);
  template <typename T1, typename T2>
  inline void chkmin(T1& x, const T2& y) {
    if (y < x) x = y;
  }
  template <typename T1, typename T2>
  inline void chkmax(T1& x, const T2& y) {
    if (x < y) x = y;
  }
  const int MAXN = 1e5 + 228;
  vector<int> g[MAXN];
  int mod = 998244353;
  int mul(int a, int b) { return 1ll * a * b % mod; }
  int add(int a, int b) {
    if (a + b >= mod) return a + b - mod;
    return a + b;
  }
  int sub(int a, int b) {
    if (a - b < 0) return a - b + mod;
    return a - b;
  }
  int binpow(int a, int b) {
    int res = 1;
    while (b) {
      if (b & 1) res = mul(res, a);
      a = mul(a, a);
      b >>= 1;
    }
    return res;
  }
  int inv(int x) { return binpow(x, mod - 2); }
  int fact[MAXN];
  int rfact[MAXN];
  int cnk(int n, int k) {
    if (n < 0 || k > n) return 0;
    return mul(fact[n], mul(rfact[k], rfact[n - k]));
  }
  int k;
  int dfs(int v, int p) {
    int cnt = 0;
    for (int i : g[v]) {
      if (i == p) continue;
      int x = dfs(i, v);
      if (x == -1) return -1;
      cnt += x;
    }
    if (cnt % k == 0) {
      if (p == v) return 0;
      return 1;
    }
    if ((cnt + 1) % k == 0 && v != p) return 0;
    return -1;
  }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cout.precision(20), cout.setf(ios::fixed);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      for (int i = 1; i <= n; ++i) g[i].clear();
      for (int i = 0; i < n - 1; ++i) {
        int a, b;
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
      }
      vector<int> dd;
      vector<int> ans(n + 1);
      ans[1] = binpow(2, n - 1);
      int x = n - 1;
      for (int i = 1; i * i <= x; ++i) {
        if (x % i == 0) {
          dd.push_back(i);
          dd.push_back(x / i);
        }
      }
      for (int i : dd) {
        if (i > 1) {
          k = i;
          ans[i] = (dfs(1, 1) == 0);
        }
      }
      for (int i = n; i >= 1; --i) {
        for (int j = i * 2; j <= n; j += i) {
          ans[i] = sub(ans[i], ans[j]);
        }
      }
      for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
      cout << '\n';
    }
  }
  