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>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  const int N = 3e5 + 9, mod = 998244353;
  int n, m, k;
  set<int> nxt[N], prv[N];
  bool bad[N];
  void dfsbads(int node) {
    if (bad[node]) return;
    bad[node] = true;
    for (auto i : nxt[node]) dfsbads(i);
    for (auto i : prv[node]) dfsbads(i);
  }
  int dfsChain(int node) {
    if (nxt[node].empty()) return 1;
    assert(nxt[node].size() == 1);
    return 1 + dfsChain(*nxt[node].begin());
  }
  int freq[N];
  int dp[N];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m >> k;
    while (n--) {
      int c, x;
      cin >> c >> x;
      for (int i = 1; i < c; i++) {
        int y;
        cin >> y;
        nxt[x].insert(y);
        prv[y].insert(x);
        x = y;
      }
    }
    for (int i = 1; i <= k; i++) {
      if (nxt[i].size() > 1 || prv[i].size() > 1) dfsbads(i);
    }
    for (int i = 1; i <= k; i++) {
      if (bad[i] || prv[i].size() > 0) continue;
      freq[dfsChain(i)]++;
    }
    vector<pair<long long, long long> > v;
    for (int i = 1; i <= m; i++) {
      if (freq[i]) {
        v.push_back({i, freq[i]});
      }
    }
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (auto j : v) {
        if (j.first > i) break;
        dp[i] = (dp[i] + 1LL * j.second * dp[i - j.first]) % mod;
      }
    }
    cout << dp[m];
  }
  