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;
  int read() {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x / 10);
    putchar(x % 10 + '0');
  }
  template <typename T>
  void print(T x, char let) {
    print(x), putchar(let);
  }
  const int N = 3e5 + 5;
  const int MOD = 998244353;
  int n, m, k, vis[N], in[N], out[N], d[N], a[N], cnt, vis1[N];
  int b[N], e[N], cnt2, vis2[N], dp[N];
  vector<int> G[N];
  bool dfs(int u, int idx) {
    vis[u] = 1;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (!vis[v] && in[v] == 1 && out[v] <= 1) {
        bool b = dfs(v, idx);
        if (b) d[u] = idx;
        return b;
      } else {
        return false;
      }
    }
    d[u] = idx;
    return true;
  }
  int main() {
    n = read();
    m = read();
    k = read();
    int x, y;
    for (int i = 1; i <= n; i++) {
      x = read();
      int l;
      for (int j = 1; j <= x; j++) {
        y = read();
        if (j != 1) {
          G[l].push_back(y);
        }
        l = y;
      }
    }
    for (int i = 1; i <= k; i++) {
      sort(G[i].begin(), G[i].end());
      vector<int>::iterator x = unique(G[i].begin(), G[i].end());
      G[i].erase(x, G[i].end());
      for (int j = 0; j < G[i].size(); j++) {
        int v = G[i][j];
        in[v]++;
        out[i]++;
      }
    }
    int tot = 0;
    for (int i = 1; i <= k; i++) {
      if (!vis[i] && in[i] == 0 && out[i] == 1) {
        dfs(i, ++tot);
      }
      if (in[i] == 0 && out[i] == 0) {
        d[i] = ++tot;
      }
    }
    for (int i = 1; i <= k; i++) {
      if (d[i]) {
        if (vis1[d[i]] == 0) {
          ++cnt;
          a[cnt] = 1;
          vis1[d[i]] = cnt;
        } else {
          a[vis1[d[i]]]++;
        }
      }
    }
    for (int i = 1; i <= cnt; i++) {
      if (vis2[a[i]] == 0) {
        ++cnt2;
        b[cnt2] = a[i];
        e[cnt2] = 1;
        vis2[a[i]] = cnt2;
      } else {
        e[vis2[a[i]]]++;
      }
    }
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= cnt2; j++) {
        if (i >= b[j]) {
          dp[i] = (dp[i] + 1ll * dp[i - b[j]] * e[j] % MOD) % MOD;
        }
      }
    }
    cout << dp[m] << endl;
    return 0;
  }
  