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;
  auto rng = bind(uniform_int_distribution<int>(1, 1000), mt19937(time(0)));
  template <class T>
  bool umin(T &a, const T &b) {
    return (a > b ? a = b, 1 : 0);
  }
  template <class T>
  bool umax(T &a, const T &b) {
    return (a < b ? a = b, 1 : 0);
  }
  const int M = 998244353;
  void add(int &a, int b) {
    a += b;
    if (a >= M)
      a -= M;
    else if (a < 0)
      a += M;
  }
  int mult(int a, int b) { return 1ll * a * b % M; }
  int binpow(int a, int b) {
    int ans = 1;
    while (b) {
      if (b & 1) ans = mult(ans, a);
      b >>= 1;
      a = mult(a, a);
    }
    return ans;
  }
  const int N = 3e5 + 1;
  bool used[N];
  int lft[N], rgt[N];
  bool isr[N];
  vector<int> pos[N];
  bool bad[N];
  vector<int> lengs;
  int ways[N];
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int>> vc(n, vector<int>());
    for (int i = 0; i < n; i++) {
      int c;
      cin >> c;
      vector<int> a(c);
      for (int j = 0; j < c; j++) {
        cin >> a[j], used[a[j]] = 1;
        if ((int)pos[a[j]].size() && pos[a[j]].back() == i) {
          bad[a[j]] = 1;
        }
        pos[a[j]].push_back(i);
      }
      vc[i] = a;
      for (int j = 0; j < c; j++) {
        if (j > 0) {
          if (!lft[a[j]] || lft[a[j]] == a[j - 1]) {
            lft[a[j]] = a[j - 1];
          } else
            bad[a[j]] = 1;
        }
        if (j + 1 < c) {
          if (!rgt[a[j]] || rgt[a[j]] == a[j + 1]) {
            rgt[a[j]] = a[j + 1];
            continue;
          } else
            bad[a[j]] = 1;
        }
      }
    }
    queue<int> q;
    for (int i = 1; i <= k; i++) {
      if (bad[i]) q.push(i);
    }
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      for (auto &j : pos[x]) {
        if (isr[j]) continue;
        isr[j] = 1;
        for (auto &v : vc[j]) {
          if (!bad[v]) {
            bad[v] = 1;
            q.push(v);
          }
        }
      }
    }
    for (int i = 1; i <= k; i++) {
      if (bad[i]) continue;
      if (!lft[i]) {
        int v = i;
        int cnt = 0;
        while (v) {
          cnt++;
          v = rgt[v];
        }
        lengs.push_back(cnt);
        ways[cnt]++;
      }
    }
    if (!(int)lengs.size()) {
      cout << 0;
      return 0;
    }
    vector<int> dp(m + 1, 0);
    sort(lengs.begin(), lengs.end());
    lengs.erase(unique(lengs.begin(), lengs.end()), lengs.end());
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (auto &j : lengs) {
        if (i >= j) add(dp[i], mult(ways[j], dp[i - j]));
      }
    }
    cout << dp[m];
    return 0;
  }
  