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;
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, m, k;
    cin >> n >> m >> k;
    set<int> v[k + 1];
    vector<int> in(k + 1, 0);
    vector<int> out(k + 1, 0);
    auto add_edge = [&](int a, int b) {
      if (v[a].find(b) != v[a].end()) return;
      v[a].insert(b);
      in[b]++;
      out[a]++;
    };
    for (int i = 1; i <= n; i++) {
      int c;
      cin >> c;
      int s;
      cin >> s;
      for (int j = 2; j <= c; j++) {
        int a;
        cin >> a;
        add_edge(s, a);
        s = a;
      }
    }
    vector<bool> vis(k + 1, 0);
    vector<int> cnt(m + 1, 0);
    for (int i = 1; i <= k; i++) {
      if (vis[i] || in[i] >= 1 || out[i] > 1) continue;
      bool ok = 1;
      int a = i;
      int len = 0;
      while (1) {
        ok &= (in[a] <= 1 && out[a] <= 1 && !vis[a]);
        if (!ok) break;
        vis[a] = 1;
        len++;
        if (out[a] == 0) break;
        a = (*v[a].begin());
      }
      if (ok && len <= m) cnt[len]++;
    }
    vector<array<int, 2>> opt;
    for (int i = 1; i <= m; i++)
      if (cnt[i] > 0) opt.push_back({i, cnt[i]});
    const long long mod = 998244353;
    auto add = [&](long long a, long long b) -> long long {
      return (a + b) % mod;
    };
    auto mul = [&](long long a, long long b) -> long long {
      return (a * b) % mod;
    };
    vector<long long> dp(m + 1, 0);
    dp[0] = 1;
    for (int i = 0; i <= m; i++) {
      for (auto [c, o] : opt)
        if (i + c <= m) dp[i + c] = add(dp[i + c], mul(o, dp[i]));
    }
    cout << dp[m] << "\n";
    return 0;
  }
  