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;
  const int LOG2 = 18;
  const int MOD = 998244353;
  const long long oo = 1e18;
  set<int> g[N];
  vector<int> g2[N];
  int in[N], seen[N], sz, bad[N], val[N];
  void mark_bad(int u) {
    bad[u] = 1;
    for (int v : g2[u])
      if (!bad[v]) mark_bad(v);
  }
  int go(int u) {
    int ok = 1;
    sz++;
    seen[u] = 1;
    ok &= g[u].size() <= 1;
    for (int v : g[u]) {
      if (!seen[v])
        ok &= go(v);
      else
        ok = 0;
    }
    return ok;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, m, k;
    cin >> n >> m >> k;
    for (int it = 0; it < n; it++) {
      int len;
      cin >> len;
      int last = -1;
      while (len--) {
        int x;
        cin >> x;
        if (last != -1) {
          in[x]++;
          g[last].insert(x);
          g2[last].push_back(x);
          g2[x].push_back(last);
        }
        last = x;
      }
    }
    map<int, int> by_sz;
    for (int i = 1; i <= k; i++) {
      if (in[i] == 0 && !bad[i]) {
        sz = 0;
        if (!go(i)) mark_bad(i);
        val[i] = sz;
      }
    }
    for (int i = 1; i <= k; i++) {
      if (!seen[i]) {
        if (!go(i)) mark_bad(i);
      }
    }
    for (int i = 1; i <= k; i++) {
      if (in[i] == 0 && !bad[i]) {
        by_sz[val[i]]++;
      }
    }
    vector<int> dp(2 * N + 2);
    dp[m] = 1;
    for (int i = m - 1; i >= 0; i--) {
      for (auto e : by_sz) {
        dp[i] += 1ll * dp[i + e.first] * e.second % MOD;
        if (dp[i] >= MOD) dp[i] -= MOD;
      }
    }
    cout << dp[0] << '\n';
  }
  