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 = 3e5 + 10, MOD = 998244353;
  struct Graph {
    vector<pair<int, long long> > son[N];
    int in[N];
    void add(int x, int y, long long w) {
      son[x].push_back(make_pair(y, w));
      in[y]++;
    }
    void init(int MAXN) {
      for (int i = 0; i <= MAXN; i++) son[i].clear(), in[i] = 0;
    }
  } gr;
  int n, m, k;
  set<int> sson[N];
  void solve() {
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++) {
      int c;
      cin >> c;
      int lst;
      for (int j = 1; j <= c; j++) {
        int x;
        cin >> x;
        if (j > 1 && !sson[lst].count(x)) {
          gr.add(lst, x, 1);
          sson[lst].insert(x);
        }
        lst = x;
      }
    }
    vector<bool> vis(k + 1);
    vector<int> cnt(k + 1);
    function<void(int, int)> dfs = [&](int u, int dep) {
      if (vis[u]) return;
      vis[u] = 1;
      if (gr.in[u] > 1) return;
      if (gr.son[u].size() > 1) return;
      if (gr.son[u].size() == 0) {
        cnt[dep]++;
        return;
      }
      dfs(gr.son[u][0].first, dep + 1);
      return;
    };
    for (int i = 1; i <= k; i++) {
      if (gr.in[i] == 0) dfs(i, 1);
    }
    vector<int> v;
    for (int i = 1; i <= k; i++)
      if (cnt[i]) {
        v.push_back(i);
      }
    vector<long long> dp(m + 1);
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (auto j : v) {
        if (i >= j) dp[i] = (dp[i] + dp[i - j] * cnt[j]) % MOD;
      }
    }
    cout << dp[m];
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int ttt = 1;
    for (int i = 1; i <= ttt; i++) solve();
    return 0;
  }
  