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 long long N = 3 * 1e5 + 100;
  long long n, m, k, in[N], out[N], w, vi[N], dp[N], t[N];
  vector<long long> e[N], p[N];
  vector<pair<long long, long long> > A;
  unordered_map<long long, long long> mp[N];
  inline void add(long long &a, long long b) {
    a = ((a + b >= 998244353) ? a + b - 998244353 : a + b);
  }
  inline void del(long long &a, long long b) {
    a = ((a - b < 0) ? a - b + 998244353 : a - b);
  }
  inline void mul(long long &a, long long b) { a = a * b % 998244353; }
  inline long long m_pow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) mul(ans, a);
      b >>= 1;
      mul(a, a);
    }
    return ans;
  }
  inline long long read() {
    long long f = 1, x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9') {
      x = x * 10 + s - '0';
      s = getchar();
    }
    return x * f;
  }
  void dfs(long long x) {
    p[w].push_back(x);
    vi[x] = 1;
    for (long long u : e[x])
      if (!vi[u]) dfs(u);
  }
  signed main() {
    n = read();
    m = read();
    k = read();
    for (long long i = 1; i <= n; i++) {
      long long c = read(), last = read();
      for (long long i = 2; i <= c; i++) {
        long long now = read();
        if (!mp[now].count(last)) {
          mp[now][last] = 1;
          in[now]++;
          out[last]++;
        }
        e[now].push_back(last);
        e[last].push_back(now);
        last = now;
      }
    }
    for (long long i = 1; i <= k; i++)
      if (!vi[i]) w++, dfs(i);
    for (long long i = 1; i <= w; i++) {
      if ((long long)p[i].size() == 1) {
        if (in[p[i][0]] == 0 && out[p[i][0]] == 0) t[1]++;
        continue;
      }
      long long cnt = 0;
      for (long long x : p[i])
        if (in[x] == 1 && out[x] == 1) cnt++;
      if (cnt != (long long)p[i].size() - 2) continue;
      long long a = 0, b = 0;
      for (long long x : p[i])
        a += (in[x] == 1 && out[x] == 0), b += (in[x] == 0 && out[x] == 1);
      if (a != 1 || b != 1) continue;
      t[(long long)p[i].size()]++;
    }
    for (long long i = 1; i <= m; i++)
      if (t[i] > 0) A.push_back(make_pair(i, t[i]));
    dp[0] = 1;
    for (long long i = 1; i <= m; i++)
      for (auto it : A)
        if (i >= it.first) add(dp[i], it.second * dp[i - it.first] % 998244353);
    printf("%lld\n", dp[m]);
  }
  