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 N = 2000100;
  const long long mod = 998244353;
  const long long inf = 2000000000000000000;
  vector<int> a[N];
  bool ban[N];
  set<int> g[N];
  set<int> in[N];
  set<int> out[N];
  bool used[N];
  vector<int> order;
  void dfs(int v) {
    order.push_back(v);
    used[v] = true;
    for (int to : g[v]) {
      if (!used[to]) dfs(to);
    }
  }
  long long dp[N];
  void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++) {
      int len;
      cin >> len;
      a[i].resize(len, 0);
      set<int> st;
      for (int &j : a[i]) cin >> j, st.insert(j);
      if (st.size() < len) {
        for (int x : st) ban[x] = true;
      } else {
        for (int j = 1; j < a[i].size(); j++) {
          g[a[i][j - 1]].insert(a[i][j]);
          g[a[i][j]].insert(a[i][j - 1]);
          out[a[i][j - 1]].insert(a[i][j]);
          in[a[i][j]].insert(a[i][j - 1]);
        }
      }
    }
    map<int, int> block;
    for (int i = 1; i <= k; i++) {
      if (used[i]) continue;
      dfs(i);
      bool OK = true;
      if (order.size() > 1) {
        int s = 0, f = 0, mid = 0;
        for (int j : order) {
          if (ban[j]) OK = false;
          if (in[j].size() == 1 && out[j].size() == 1)
            mid++;
          else if (in[j].size() == 1 && out[j].size() == 0)
            f++;
          else if (in[j].size() == 0 && out[j].size() == 1)
            s++;
        }
        if (s != 1 || f != 1 || mid + 2 != order.size()) OK = false;
      } else
        OK = (!ban[i]);
      if (OK) block[order.size()]++;
      order.clear();
    }
    vector<pair<int, int> > v;
    for (auto cur : block) v.push_back(cur);
    dp[0] = 1ll;
    for (int i = 1; i <= m; i++) {
      dp[i] = 0ll;
      for (auto cur : v) {
        if (i >= cur.first) {
          dp[i] += dp[i - cur.first] * cur.second;
          dp[i] %= mod;
        }
      }
    }
    cout << dp[m] << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tt = 1;
    while (tt--) {
      solve();
    }
    return 0;
  }
  