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 maxn = 300005, mod = 998244353;
  int n, m, k, mx;
  int in[maxn], out[maxn], f[maxn], tot[maxn], q[maxn], tmp[maxn];
  vector<int> t, v[maxn], g[maxn];
  map<int, int> mp[maxn];
  void dfs(int x, int d) {
    if (in[x] > 1 || out[x] > 1) return;
    if (out[x] == 0) tot[d]++, mx = max(mx, d);
    for (int i = 0; i < g[x].size(); i++) dfs(g[x][i], d + 1);
  }
  int main() {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1, c; i <= n; i++) {
      scanf("%d", &c), v[i].resize(c);
      for (int j = 0; j < c; j++) {
        scanf("%d", &v[i][j]);
        if (j > 0 && mp[v[i][j - 1]][v[i][j]] == 0)
          mp[v[i][j - 1]][v[i][j]] = 1, g[v[i][j - 1]].push_back(v[i][j]),
                      out[v[i][j - 1]]++, in[v[i][j]]++;
      }
    }
    for (int i = 1; i <= k; i++)
      if (in[i] == 0) dfs(i, 1);
    f[0] = 1;
    for (int i = 1; i <= mx; i++)
      if (tot[i]) t.push_back(i);
    for (int i = 1; i <= m; i++)
      for (int j = 0; j < t.size(); j++)
        if (i >= t[j]) f[i] = (f[i] + 1ll * f[i - t[j]] * tot[t[j]]) % mod;
    printf("%d\n", f[m]);
    return 0;
  }
  