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 MAX = 3e5 + 5;
  const int MOD = 998244353;
  vector<int> bio(MAX);
  vector<int> degree(MAX);
  vector<int> edges[MAX];
  int add(int x, int y) {
    x += y;
    while (x >= MOD) {
      x -= MOD;
    }
    while (x < 0) {
      x += MOD;
    }
    return x;
  }
  int mul(int x, int y) { return (x * 1ll * y) % MOD; }
  int dfs(int a) {
    if (degree[a] > 1 || edges[a].size() > 1) return 0;
    if (edges[a].size() == 0) return 1;
    int z = dfs(edges[a][0]);
    if (z == 0)
      return 0;
    else
      return z + 1;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 0; i < n; i++) {
      int c, a, b;
      cin >> c >> a;
      c--;
      while (c--) {
        cin >> b;
        if (edges[a].size() == 0 || edges[a][0] != b) {
          edges[a].push_back(b);
          degree[b]++;
        }
        a = b;
      }
    }
    map<int, int> mp;
    for (int i = 1; i <= k; i++) {
      if (degree[i] == 0) {
        int z = dfs(i);
        if (z) mp[z]++;
      }
    }
    vector<int> dp(m + 1);
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (auto e : mp) {
        if (i < e.first) continue;
        dp[i] = add(dp[i], mul(dp[i - e.first], e.second));
      }
    }
    cout << dp[m] << "\n";
    return 0;
  }
  