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>
  const long long mod = 998244353;
  int main() {
    std::ios::sync_with_stdio(false);
    std::cout.tie(0);
    int n, m, k;
    std::cin >> n >> m >> k;
    std::vector<std::vector<int> > a(n, std::vector<int>(0));
    std::set<int> ban;
    std::vector<std::vector<int> > vec;
    for (int i = 0; i < n; i++) {
      int c, x;
      std::cin >> c;
      while (c--) {
        std::cin >> x;
        x--;
        a[i].emplace_back(x);
      }
      std::map<int, int> mp;
      for (auto x : a[i]) mp[x]++;
      bool flag = true;
      for (auto p : mp)
        if (p.second >= 2) {
          ban.insert(p.first);
          flag = false;
        }
      if (not flag) {
        for (auto x : a[i]) ban.insert(x);
      }
    }
    std::vector<std::set<int> > g(k), rg(k);
    for (auto v : a) {
      for (int i = 0; i + 1 < v.size(); i++) {
        g[v[i]].insert(v[i + 1]);
        rg[v[i + 1]].insert(v[i]);
      }
    }
    std::vector<int> in(k, 0), out(k, 0);
    for (int i = 0; i < k; i++) {
      out[i] = g[i].size();
      in[i] = rg[i].size();
    }
    for (int i = 0; i < k; i++) {
      if (in[i] >= 2 || out[i] >= 2) ban.insert(i);
    }
    std::function<int(int)> dfs = [&](int x) {
      if (ban.count(x)) return -(int)1e9;
      if (out[x] == 0) return 1;
      if (out[x] == 1 && in[*g[x].begin()] == 1) {
        return 1 + dfs(*g[x].begin());
      }
      return -(int)1e9;
    };
    std::vector<int> F(k + 1, 0);
    for (int i = 0; i < k; i++)
      if (!ban.count(i)) {
        if (in[i] == 0) {
          int len = dfs(i);
          if (len > 0) F[len]++;
        }
      }
    std::vector<std::pair<int, int> > key;
    for (int i = 0; i <= k; i++)
      if (F[i] > 0) {
        key.emplace_back(i, F[i]);
      }
    std::vector<int> dp(m + 1, 0);
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (auto [len, w] : key) {
        assert(len >= 1 && len <= k);
        if (len <= i) {
          dp[i] = (dp[i] + 1ll * w * dp[i - len] % mod) % mod;
        }
      }
    }
    std::cout << dp[m] << "\n";
  }
  