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;
  template <typename T>
  inline void read(T &f) {
    f = 0;
    T fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') fu = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      f = (f << 3) + (f << 1) + (c & 15);
      c = getchar();
    }
    f *= fu;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10)
      putchar(x + 48);
    else
      print(x / 10), putchar(x % 10 + 48);
  }
  template <typename T>
  void print(T x, char t) {
    print(x);
    putchar(t);
  }
  const int md = 998244353;
  inline int add(int x, int y) {
    if (x + y >= md) return x + y - md;
    return x + y;
  }
  inline void addx(int &x, int y) {
    x += y;
    if (x >= md) x -= md;
  }
  inline int sub(int x, int y) {
    if (x < y) return x - y + md;
    return x - y;
  }
  inline void subx(int &x, int y) {
    x -= y;
    if (x < 0) x += md;
  }
  inline int mul(int x, int y) { return 1ull * x * y % md; }
  inline int fpow(int x, int y) {
    int ans = 1;
    while (y) {
      if (y & 1) ans = mul(ans, x);
      y >>= 1;
      x = mul(x, x);
    }
    return ans;
  }
  const int N = 3e5 + 5;
  pair<int, int> g[N];
  int nxt[N], pre[N], a[N], ban[N], vis[N], t[N], dp[N];
  int n, m, k, len, flag;
  void link(int u, int v) {
    if (nxt[u] && nxt[u] != v) {
      flag = 1;
      return;
    }
    if (pre[v] && pre[v] != u) {
      flag = 1;
      return;
    }
    nxt[u] = v;
    pre[v] = u;
  }
  int main() {
    read(n);
    read(m);
    read(k);
    for (int i = 1; i <= n; i++) {
      int len;
      read(len);
      flag = 0;
      for (int j = 1; j <= len; j++) {
        read(a[j]);
        if (j > 1) link(a[j - 1], a[j]);
      }
      if (flag) {
        for (int j = 1; j <= len; j++) {
          ban[a[j]] = 1;
        }
      }
    }
    for (int i = 1; i <= k; i++) {
      if (ban[i]) continue;
      if (!pre[i]) {
        int tmp = i, cando = 1;
        while (tmp) {
          if (ban[tmp]) cando = 0;
          tmp = nxt[tmp];
        }
        if (cando) {
          int qwq = 0;
          tmp = i;
          while (tmp) {
            ++qwq;
            vis[tmp] = 1;
            tmp = nxt[tmp];
          }
          ++t[qwq];
        }
      }
    }
    for (int i = 1; i <= k; i++) {
      if (t[i]) g[++len] = make_pair(i, t[i]);
    }
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= len; j++) {
        if (g[j].first > i) break;
        dp[i] = add(dp[i], mul(dp[i - g[j].first], g[j].second));
      }
    }
    print(dp[m], '\n');
    return 0;
  }
  