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 c = 300005;
  int n, m, k, db[c], el[c], kov[c], ki[c];
  vector<int> sz[c];
  bool rossz[c], v[c];
  vector<pair<int, long long> > lep;
  long long dp[c], mod = 998244353;
  void dfs(int a) {
    v[a] = true;
    rossz[a] = 1;
    for (auto x : sz[a]) {
      if (!v[x]) {
        dfs(x);
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++) {
      int h, pr;
      cin >> h;
      cin >> pr;
      ki[pr] = i;
      for (int j = 1; j < h; j++) {
        int x;
        cin >> x;
        sz[x].push_back(pr);
        sz[pr].push_back(x);
        if (ki[x] == i) {
          rossz[x] = 1;
        }
        ki[x] = i;
        if (kov[pr] && kov[pr] != x) {
          rossz[pr] = 1;
        }
        if (el[x] && el[x] != pr) {
          rossz[x] = 1;
        }
        kov[pr] = x;
        el[x] = pr;
        pr = x;
      }
    }
    for (int i = 1; i <= k; i++) {
      if (rossz[i] && !v[i]) {
        dfs(i);
      }
    }
    for (int i = 1; i <= k; i++) {
      if (!rossz[i] && !el[i]) {
        int len = 1, pos = i;
        while (kov[pos]) {
          pos = kov[pos];
          len++;
        }
        db[len]++;
      }
    }
    for (int i = 1; i <= k; i++) {
      if (db[i]) {
        lep.push_back({i, db[i]});
      }
    }
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (auto x : lep) {
        if (i >= x.first) {
          dp[i] += dp[i - x.first] * x.second;
        }
      }
      dp[i] %= mod;
    }
    cout << dp[m] << "\n";
    return 0;
  }
  