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 T>
  void read(T &x) {
    T f = 1;
    x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9') {
      x = (x << 3) + (x << 1) + (s ^ '0');
      s = getchar();
    }
    x *= f;
  }
  template <typename T>
  void write(T x, char s = '\n') {
    if (!x) {
      putchar('0');
      putchar(s);
      return;
    }
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    T t = 0, tmp[25] = {};
    while (x) tmp[t++] = x % 10, x /= 10;
    while (t-- > 0) putchar(tmp[t] + '0');
    putchar(s);
  }
  const int MAXN = 1e5 + 5;
  const int mod = 998244353;
  long long ans[MAXN], dp[MAXN][2], deg[MAXN];
  int head[MAXN], to[MAXN << 1], nxt[MAXN << 1], cnt, n;
  void add(int u, int v) {
    nxt[++cnt] = head[u];
    head[u] = cnt;
    to[cnt] = v;
  }
  long long tmp[MAXN];
  void dfs(int x, int fa, long long k) {
    dp[x][0] = dp[x][1] = 0;
    for (int i = head[x]; i; i = nxt[i]) {
      int v = to[i];
      if (v == fa) continue;
      dfs(v, x, k);
    }
    if (k > 1) {
      if (deg[x] < k) {
        dp[x][1] = 0;
        dp[x][0] = 1;
        for (int i = head[x]; i; i = nxt[i]) {
          int v = to[i];
          if (v == fa) continue;
          dp[x][0] *= dp[v][1];
        }
      } else {
        long long zer = 0, one = 0, nei = 0;
        for (int i = head[x]; i; i = nxt[i]) {
          int v = to[i];
          if (v == fa) continue;
          if (dp[v][1])
            one++;
          else if (dp[v][0])
            zer++;
          else
            nei++;
        }
        if (nei)
          dp[x][0] = dp[x][1] = 0;
        else {
          if (zer % k == 0) dp[x][0] = 1;
          if ((zer + 1) % k == 0) dp[x][1] = 1;
        }
      }
    } else {
      for (int i = 0; i < k; ++i) tmp[i] = 0;
      tmp[0] = 1;
      for (int i = head[x]; i; i = nxt[i]) {
        int v = to[i];
        if (v == fa) continue;
        long long las = tmp[k - 1];
        for (int j = k - 1; j > 0; --j)
          tmp[j] = (tmp[j] * dp[v][1] + tmp[j - 1] * dp[v][0]) % mod;
        tmp[0] = (tmp[0] * dp[v][1] + las * dp[v][0]) % mod;
      }
      dp[x][0] = tmp[0], dp[x][1] = tmp[k - 1];
    }
  }
  long long calc(long long x) {
    dfs(1, 0, x);
    return dp[1][0];
  }
  int main() {
    int T;
    read(T);
    while (T-- > 0) {
      read(n);
      for (int i = 1; i < n; ++i) {
        int u, v;
        read(u), read(v);
        add(u, v), add(v, u);
        deg[u]++, deg[v]++;
      }
      for (int i = 1; i * i <= n - 1; ++i) {
        if ((n - 1) % i) continue;
        ans[i] = calc(i);
        if ((n - 1) / i != i) ans[(n - 1) / i] = calc((n - 1) / i);
      }
      for (int i = (n - 1); i >= 1; --i)
        if (ans[i]) {
          for (int j = i * 2; j <= (n - 1); j += i)
            ans[i] = (ans[i] - ans[j] + mod) % mod;
        }
      for (int i = 1; i <= n; ++i)
        write(ans[i], ' '), ans[i] = 0, deg[i] = 0, head[i] = 0;
      cnt = 0;
      puts("");
    }
    return 0;
  }
  