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 = 300010;
  const int mod = 998244353;
  const int inf = 2147483647;
  long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') x = x * 10ll + ch - '0', ch = getchar();
    return x * f;
  }
  void up(int& x, int y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  int n, m, k, L[N], R[N], a[N], b[N], cnt[N], f[N];
  bool v[N], del[N];
  vector<pair<int, int> > h;
  int main() {
    n = read(), m = read(), k = read();
    for (int i = 1; i <= n; i++) {
      int l = read();
      for (int j = 1; j <= l; j++) a[j] = read(), cnt[a[j]]++;
      bool flag = false;
      for (int j = 1; j <= l; j++)
        if (cnt[a[j]] > 1) {
          flag = true;
          break;
        }
      for (int j = 1; j <= l; j++) cnt[a[j]]--;
      for (int j = 2; j <= l; j++) {
        if (R[a[j - 1]] && R[a[j - 1]] != a[j]) flag = true;
        R[a[j - 1]] = a[j];
        v[a[j]] = true;
      }
      for (int j = 1; j < l; j++) {
        if (L[a[j + 1]] && L[a[j + 1]] != a[j]) flag = true;
        L[a[j + 1]] = a[j];
      }
      if (flag) {
        for (int j = 1; j <= l; j++) del[a[j]] = true;
      }
    }
    for (int i = 1; i <= k; i++)
      if (!v[i]) {
        int cnt = 0, t = i;
        bool tmp = false;
        while (t) {
          tmp |= del[t];
          if (tmp) break;
          cnt++;
          t = R[t];
        }
        if (!tmp) b[cnt]++;
      }
    for (int i = 1; i <= m; i++)
      if (b[i]) h.push_back(make_pair(i, b[i]));
    f[0] = 1;
    for (int i = 1; i <= m; i++)
      for (pair<int, int> j : h)
        if (j.first <= i) up(f[i], (long long)f[i - j.first] * j.second % mod);
    printf("%d\n", f[m]);
    return 0;
  }
  