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 + 10, mod = 1e9 + 7;
  int cnt = 0;
  int s[N], dp[N];
  char g[N];
  string str;
  vector<int> vec;
  long long su(long long a, long long b) {
    a += b;
    return (a >= 998244353) ? a - 998244353 : a;
  }
  int i, j, k, n, m, t, a[300500], in[300500];
  long long f[300500], num[300500];
  map<pair<int, int>, int> mp;
  vector<int> v[300050], v1;
  void dfs(int first, int dep) {
    if (v[first].size() > 1 || in[first] > 1) return;
    if (!v[first].size()) {
      if (!num[dep]) v1.push_back(dep);
      num[dep]++;
      return;
    }
    for (auto i : v[first]) dfs(i, dep + 1);
  }
  int main() {
    f[0] = 1;
    scanf("%d%d%d", &t, &m, &k);
    while (t--) {
      scanf("%d", &n);
      for (i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        if (i > 1) {
          if (!mp[{a[i - 1], a[i]}]) {
            v[a[i - 1]].push_back(a[i]), in[a[i]]++;
            mp[{a[i - 1], a[i]}] = 1;
          }
        }
      }
    }
    for (i = 1; i <= k; i++) {
      if (!in[i]) dfs(i, 1);
    }
    for (i = 1; i <= m; i++) {
      for (auto j : v1) {
        if (j <= i) f[i] = su(f[i], f[i - j] * num[j] % 998244353);
      }
    }
    printf("%lld", f[m]);
  }
  