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;
  inline int read() {
    int n = 0, f = 1, ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      n = n * 10 + ch - '0';
      ch = getchar();
    }
    return n * f;
  }
  char str[200005];
  int a[200005];
  vector<int> v[200005];
  int k;
  int dp[200005][2], rd[200005];
  int da[200005], jda[200005];
  void dfs(int x, int f) {
    if (rd[x] < k) {
      dp[x][0] = 1;
      for (int i = 0; i < v[x].size(); i++) {
        if (v[x][i] == f) continue;
        dfs(v[x][i], x);
        dp[x][0] = 1LL * dp[x][0] * dp[v[x][i]][1] % 998244353;
      }
      dp[x][1] = 0;
      return;
    }
    int cj = 1, sl = 0;
    for (int i = 0; i < v[x].size(); i++) {
      if (v[x][i] == f) continue;
      dfs(v[x][i], x);
      if (dp[v[x][i]][1] == 0) {
        cj = 1LL * dp[v[x][i]][0] * cj % 998244353;
        sl++;
      }
    }
    for (int i = 0; i < k; i++) da[i] = 0;
    da[sl % k] = cj;
    for (int i = 0; i < v[x].size(); i++) {
      if (v[x][i] == f) continue;
      if (dp[v[x][i]][1] == 0) continue;
      for (int j = 0; j < k; j++) jda[j] = da[j];
      for (int j = 0; j < k; j++) {
        da[j] = 1LL * jda[j] * dp[v[x][i]][1] % 998244353;
        if (j == 0)
          da[j] =
              (da[j] + 1LL * jda[k - 1] * dp[v[x][i]][0] % 998244353) % 998244353;
        else
          da[j] =
              (da[j] + 1LL * jda[j - 1] * dp[v[x][i]][0] % 998244353) % 998244353;
      }
    }
    dp[x][0] = da[0];
    dp[x][1] = da[k - 1];
  }
  int tans[200005];
  int main() {
    int t, n, x, y;
    t = read();
    for (int greg = 1; greg <= t; greg++) {
      n = read();
      for (int i = 1; i <= n - 1; i++) {
        x = read();
        y = read();
        v[x].push_back(y);
        v[y].push_back(x);
        rd[x]++;
        rd[y]++;
      }
      for (int i = 1; i <= n; i++) {
        if ((n - 1) % i == 0) {
          k = i;
          dfs(1, 0);
          tans[i] = dp[1][0];
        } else
          tans[i] = 0;
      }
      for (int i = n; i >= 1; i--) {
        for (int j = 2 * i; j <= n; j += i)
          tans[i] = (tans[i] + 998244353 - tans[j]) % 998244353;
      }
      for (int i = 1; i <= n; i++) printf("%d ", tans[i]);
      printf("\n");
      for (int i = 1; i <= n; i++) {
        v[i].clear();
        dp[i][0] = dp[i][1] = 0;
        rd[i] = 0;
      }
    }
    return 0;
  }
  