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 long long MOD = 998244353;
  long long n, m, t, nn = 0, vis[300010], cnt[300010], dp[300010], extr = 0;
  vector<long long> a[300010], g[300010], gg[300010], tmpa[300010], pth;
  vector<pair<long long, long long> > v;
  set<long long> st;
  bool bada[300010], badc[300010];
  map<long long, long long> s;
  void dfs(long long pos, long long ty) {
    if ((ty == 0 && bada[pos]) || (ty == 1 && badc[pos])) return;
    if (ty == 0) {
      bada[pos] = true;
      for (int i = 0; i < a[pos].size(); ++i) dfs(a[pos][i], 1);
    } else {
      badc[pos] = true;
      for (int i = 0; i < g[pos].size(); ++i) dfs(g[pos][i], 0);
    }
  }
  bool dfs2(long long node, long long strPos, long long sPos) {
    long long stPos = sPos - strPos;
    if (bada[node]) return false;
    if (vis[node] > -1e18) return vis[node] == stPos;
    vis[node] = stPos;
    pth.push_back(node);
    vector<long long> trans;
    trans.clear();
    for (int i = 0; i < a[node].size(); ++i) {
      if (s.find(stPos + i) != s.end()) {
        if (s[stPos + i] != a[node][i]) return false;
      } else {
        trans.push_back(i);
        s[stPos + i] = a[node][i];
      }
    }
    for (int i = 0; i < trans.size(); ++i) {
      long long val = a[node][trans[i]];
      for (int j = 0; j < g[val].size(); ++j)
        if (!dfs2(g[val][j], gg[val][j], stPos + trans[i])) return false;
    }
    return true;
  }
  int main() {
    cin >> n >> m >> t;
    long long x, y;
    for (int i = 0; i < n; ++i) {
      scanf("%lld", &x);
      for (int j = 0; j < x; ++j) {
        scanf("%lld", &y);
        a[i].push_back(y);
        g[y].push_back(i);
        st.insert(y);
      }
    }
    for (int i = 1; i <= t; ++i)
      if (st.find(i) == st.end()) ++extr;
    for (int i = 0; i < n; ++i)
      if (!bada[i]) {
        st.clear();
        for (int j = 0; j < a[i].size(); ++j) st.insert(a[i][j]);
        if (st.size() < a[i].size())
          for (auto it : st)
            if (!badc[it]) dfs(it, 1);
      }
    for (int i = 0; i < n; ++i)
      if (!bada[i]) tmpa[nn++] = a[i];
    n = nn;
    for (int i = 0; i < n; ++i) a[i] = tmpa[i];
    for (int i = 0; i < t + 5; ++i) g[i].clear();
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < a[i].size(); ++j)
        g[a[i][j]].push_back(i), gg[a[i][j]].push_back(j);
    for (int i = 0; i < n + 5; ++i) vis[i] = -1e18, bada[i] = false;
    for (int i = 0; i < n; ++i)
      if (vis[i] == -1e18) {
        s.clear();
        pth.clear();
        bool stat = dfs2(i, 0, 0);
        if (!stat)
          for (int j = 0; j < pth.size(); ++j) bada[pth[j]] = true;
        else
          ++cnt[(s.rbegin()->first) - (s.begin()->first) + 1];
      }
    for (int i = 1; i <= m; ++i)
      if (cnt[i] > 0) v.push_back(make_pair(i, cnt[i]));
    dp[0] = 1;
    for (int i = 1; i <= m; ++i) {
      (dp[i] += dp[i - 1] * extr) %= MOD;
      for (int j = 0; j < v.size(); ++j)
        if (i - v[j].first >= 0)
          (dp[i] += dp[i - v[j].first] * v[j].second) %= MOD;
        else
          break;
    }
    cout << dp[m] << endl;
    return 0;
  }
  