description: A subarray of array a from index l to the index r is the array [a_l, a_{l+1}, ..., a_{r}]. The number of occurrences of the array b in the array a is the number of subarrays of a such that they are equal to b.
  
  You are given n arrays A_1, A_2, ..., A_n; the elements of these arrays are integers from 1 to k. You have to build an array a consisting of m integers from 1 to k in such a way that, for every given subarray A_i, the number of occurrences of A_i in the array a is not less than the number of occurrences of each non-empty subarray of A_i in a. Note that if A_i doesn't occur in a, and no subarray of A_i occurs in a, this condition is still met for A_i.
  
  Your task is to calculate the number of different arrays a you can build, and print it modulo 998244353.
  
  Input
  
  The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 3 ⋅ 10^5) — the number of the given arrays, the desired length of the array a, and the upper bound on the values in the arrays.
  
  Then n lines follow. The i-th line represents the array A_i. The first integer in the i-th line is c_i (1 ≤ c_i ≤ m) — the number of elements in A_i; then, c_i integers from 1 to k follow — the elements of the array A_i.
  
  Additional constraint on the input: ∑_{i=1}^n c_i ≤ 3 ⋅ 10^5; i. e., the number of elements in the given arrays in total does not exceed 3 ⋅ 10^5.
  
  Output
  
  Print one integer — the number of different arrays a you can build, taken modulo 998244353.
  
  Examples
  
  Input
  
  
  2 4 3
  2 1 2
  1 3
  
  
  Output
  
  
  5
  
  
  Input
  
  
  2 4 3
  2 1 2
  3 3 2 1
  
  
  Output
  
  
  0
  
  
  Input
  
  
  1 42 1337
  2 13 31
  
  
  Output
  
  
  721234447
solution: #include <bits/stdc++.h>
  using namespace std;
  const int MAXN = 300005;
  const int Mod = 998244353;
  vector<int> pre[MAXN], nxt[MAXN];
  bool vis[MAXN];
  void dfs(int u, int las, int& ok, int& siz) {
    vis[u] = 1;
    ++siz;
    ok &= (pre[u].size() <= 1u);
    ok &= (nxt[u].size() <= 1u);
    for (auto& v : pre[u]) {
      if (v == las) continue;
      if (vis[v]) {
        ok = 0;
        continue;
      }
      dfs(v, u, ok, siz);
    }
    for (auto& v : nxt[u]) {
      if (v == las) continue;
      if (vis[v]) {
        ok = 0;
        continue;
      }
      dfs(v, u, ok, siz);
    }
  }
  int main() {
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 0; i < n; i++) {
      int c;
      scanf("%d", &c);
      vector<int> a(c);
      for (int j = 0; j < c; j++) scanf("%d", &a[j]);
      for (int j = 0; j + 1 < c; j++) {
        nxt[a[j]].push_back(a[j + 1]);
        pre[a[j + 1]].push_back(a[j]);
      }
    }
    for (int i = 1; i <= k; i++) {
      sort(nxt[i].begin(), nxt[i].end());
      nxt[i].erase(unique(nxt[i].begin(), nxt[i].end()), nxt[i].end());
      sort(pre[i].begin(), pre[i].end());
      pre[i].erase(unique(pre[i].begin(), pre[i].end()), pre[i].end());
    }
    map<int, int> mp;
    for (int i = 1; i <= k; i++) {
      if (vis[i]) continue;
      int ok = 1, siz = 0;
      dfs(i, 0, ok, siz);
      if (ok) mp[siz]++;
    }
    static int dp[MAXN];
    for (int i = (dp[0] = 1); i <= m; i++)
      for (auto& t : mp) {
        int j = t.first, c = t.second;
        if (j <= i) dp[i] = (dp[i] + 1LL * c * dp[i - j]) % Mod;
      }
    printf("%d\n", dp[m]);
    return 0;
  }
  