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;
  using LL = long long;
  constexpr LL mod = 998244353;
  void no() {
    cout << "0";
    exit(0);
  }
  int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int n, m, k;
    cin >> n >> m >> k;
    vector<set<int>> G(k + 1), H(k + 1);
    for (int i = 0; i < n; i += 1) {
      int c;
      cin >> c;
      vector<int> A(c);
      for (int& x : A) cin >> x;
      for (int j = 1; j < c; j += 1) {
        G[A[j - 1]].insert(A[j]);
        H[A[j]].insert(A[j - 1]);
      }
    }
    map<int, int> mp;
    for (int i = 1; i <= k; i += 1)
      if (H[i].empty()) {
        int x = 0;
        for (int j = i; j;) {
          x += 1;
          if (H[j].size() > 1) {
            x = -1;
            break;
          }
          if (G[j].empty()) break;
          if (G[j].size() > 1) {
            x = -1;
            break;
          }
          j = *G[j].begin();
        }
        if (x != -1) mp[x] += 1;
      }
    vector<LL> dp(m + 1);
    dp[0] = 1;
    for (int i = 1; i <= m; i += 1)
      for (auto [x, y] : mp)
        if (i >= x) dp[i] = (dp[i] + dp[i - x] * y) % mod;
    cout << dp[m];
    return 0;
  }
  