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 + 5, mod = 998244353;
  int t, vis[N], p[2 * N], n, m, K, cnt[N], dp[N], l, r, F, first[N], c[N];
  set<int> second[N];
  vector<int> a[N], V[N];
  void dfs(int u, int id) {
    if (vis[u]) return;
    vis[u] = 1;
    while (second[u].size()) {
      int x = *second[u].begin();
      for (int i = 0; i < a[x].size(); i++) second[a[x][i]].erase(x);
      int first = -1;
      for (int j = 0; j < a[x].size(); j++) {
        if (a[x][j] == u) first = j;
      }
      assert(first != -1);
      for (int j = (int)a[x].size() - 1; j >= 0; j--) {
        int d = a[x][j];
        l = min(l, id + j - first);
        r = max(r, id + j - first);
        if (l <= 0) {
          F = 1;
        } else {
          if (!p[id + j - first])
            p[id + j - first] = d;
          else if (p[id + j - first] != d) {
            F = 1;
          }
        }
        if (!vis[d]) {
          dfs(d, id + j - first);
        }
      }
    }
  }
  int main() {
    cin >> n >> m >> K;
    for (int i = 1; i <= n; i++) {
      int c;
      cin >> c;
      while (c--) {
        int x;
        cin >> x;
        a[i].push_back(x);
      }
    }
    for (int i = 1; i <= n; i++) {
      int F = 0;
      for (int j = 0; j < a[i].size(); j++) {
        second[a[i][j]].insert(i);
      }
    }
    for (int i = 1; i <= K; i++) {
      if (!vis[i] && !first[i]) {
        l = r = m + 1;
        F = 0;
        p[m + 1] = i;
        dfs(i, m + 1);
        for (int j = max(0, l); j <= r; j++) {
          if (c[p[j]] && p[j]) F = 1;
          c[p[j]]++;
        }
        for (int j = max(0, l); j <= r; j++) c[p[j]] = 0, p[j] = 0;
        if (!(r - l + 1 > m || F)) cnt[r - l + 1]++;
      }
    }
    dp[0] = 1;
    vector<pair<int, int> > x;
    for (int j = 1; j <= m; j++) {
      if (cnt[j]) x.push_back({j, cnt[j]});
    }
    for (int i = 1; i <= m; i++) {
      for (int j = 0; j < x.size(); j++)
        if (i - x[j].first >= 0)
          dp[i] =
              (dp[i] + (long long)dp[i - x[j].first] * x[j].second % mod) % mod;
    }
    cout << endl;
    cout << dp[m];
  }
  