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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target( \
      "fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  const double eps = 1e-6;
  struct inout {
    static const long long ibufl = 1 << 25;
    char in_buf[ibufl + 5], out_buf[ibufl + 5], *inf, *ouf;
    void init() {
      fread(in_buf, 1, ibufl, stdin);
      inf = in_buf;
      ouf = out_buf;
    }
    inout& operator>>(long long& a) {
      long long fh = 1;
      while (!(isdigit(*inf) || *inf == '-')) ++inf;
      if (*inf == '-') fh = -1, ++inf;
      a = 0;
      while (isdigit(*inf)) {
        a = a * 10 + *inf - '0';
        ++inf;
      }
      a *= fh;
      return *this;
    }
    inout& operator>>(char& a) {
      while (*inf == ' ' || *inf == '\n') ++inf;
      a = *inf;
      ++inf;
      return *this;
    }
    inout& operator>>(char* a) {
      while (*inf == ' ' || *inf == '\n') ++inf;
      while (!(*inf == ' ' || *inf == '\n')) {
        *a = *inf;
        ++inf;
        ++a;
      }
      *a = '\0';
      return *this;
    }
    inout& operator>>(double& a) {
      long long fh = 1;
      double s;
      while (!(isdigit(*inf) || *inf == '-')) ++inf;
      if (*inf == '-') fh = -1, ++inf;
      a = 0;
      while (isdigit(*inf)) {
        a = a * 10 + *inf - '0';
        ++inf;
      }
      if (*inf == '.') {
        s = 0.1;
        ++inf;
        while (isdigit(*inf)) {
          a += s * (*inf - '0');
          ++inf;
          s *= 0.1;
        }
      }
      a *= fh;
      return *this;
    }
    void writeint(long long x) {
      if (x / 10) writeint(x / 10);
      *ouf = x % 10 + '0';
      ++ouf;
    }
    inout& operator<<(long long a) {
      if (a < 0) {
        *ouf = '-';
        ++ouf;
        a = -a;
      }
      writeint(a);
      return *this;
    }
    static const long long sz = 2;
    inout& operator<<(char a) {
      *ouf = a;
      ++ouf;
      return *this;
    }
    inout& operator<<(char* a) {
      while (*a) {
        *ouf = *a;
        ++ouf;
        ++a;
      }
      return *this;
    }
    inout& operator<<(double a) {
      if (a < -eps) {
        *ouf = '-';
        ++ouf;
        a = -a;
      }
      writeint((long long)a);
      a -= (long long)a;
      *ouf = '.';
      ++ouf;
      for (long long i = 1; i <= sz; i++) {
        a = a * 10;
        *ouf = (long long)a + '0';
        ++ouf;
        a -= (long long)a;
      }
      return *this;
    }
    void out() { fwrite(out_buf, 1, ouf - out_buf, stdout); }
  };
  inout io;
  long long L[300005], R[300005];
  long long cnt[300005], fa[300005], siz[300005];
  bool down[300005], vis[300005];
  long long n, m, k;
  long long dp[300005];
  long long get(long long x) { return (fa[x] == x) ? x : (fa[x] = get(fa[x])); }
  bool merge(long long x, long long y) {
    x = get(x);
    y = get(y);
    if (x == y) return 0;
    fa[x] = y;
    siz[y] += siz[x];
    down[y] |= down[x];
    return 1;
  }
  signed main() {
    io.init();
    io >> n >> m >> k;
    for (long long i = 1; i <= k; i++) fa[i] = i, siz[i] = 1;
    for (long long i = 1; i <= n; i++) {
      long long len, x;
      io >> len >> x;
      for (long long j = 1; j < len; j++) {
        long long y;
        io >> y;
        if (!R[x]) {
          if (!merge(x, y)) {
            down[x] = down[y] = 1;
          }
        }
        merge(x, y);
        if (R[x] && R[x] != y) {
          down[x] = 1;
        }
        if (L[y] && L[y] != x) {
          down[y] = 1;
        }
        R[x] = y;
        L[y] = x;
        x = y;
      }
    }
    for (long long i = 1; i <= k; i++) down[get(i)] |= down[i];
    for (long long i = 1; i <= k; i++)
      if (!vis[fa[i]] && !down[fa[i]]) {
        cnt[siz[fa[i]]]++;
        vis[fa[i]] = 1;
      }
    vector<long long> v;
    for (long long i = 1; i <= m; i++)
      if (cnt[i]) {
        v.emplace_back(i);
      }
    dp[0] = 1;
    long long mod = 998244353;
    for (long long i = 1; i <= m; i++) {
      for (auto u : v)
        if (u <= i) {
          dp[i] = (1ll * dp[i - u] * cnt[u] + dp[i]) % mod;
        }
    }
    cout << dp[m];
    return 0;
  }
  