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(3)
  using namespace std;
  const int N = 4000005, mod = 998244353;
  int n, m, k, l[N], r[N], a[N], vis[N], t, mx, b[N], fl;
  vector<int> s, p;
  int qp(int a, int b) {
    int ans = 1;
    while (b) {
      if (b & 1) ans = 1ll * ans * a % mod;
      a = 1ll * a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  int rv[N], w[N];
  void ntt(vector<int>& a, int f) {
    int n = a.size();
    for (int i = 0; i < n; i++) rv[i] = (rv[i >> 1] >> 1) | ((i & 1) * (n >> 1));
    int wn = qp(f ? (mod + 1) / 3 : 3, (mod - 1) / n);
    w[0] = 1;
    for (int i = 1; i < n; i++) w[i] = 1ll * w[i - 1] * wn % mod;
    for (int i = 0; i < n; i++)
      if (i < rv[i]) swap(a[i], a[rv[i]]);
    for (int mid = 1; mid < n; mid <<= 1) {
      for (int i = 0; i < n; i += (mid << 1)) {
        for (int j = 0; j < mid; j++) {
          int x = a[i + j],
              y = 1ll * a[i + j + mid] * w[n / (mid << 1) * j] % mod;
          a[i + j] = (x + y >= mod ? x + y - mod : x + y);
          a[i + j + mid] = (x - y < 0 ? x - y + mod : x - y);
        }
      }
    }
    if (f) {
      int iv = qp(n, mod - 2);
      for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * iv % mod;
    }
  }
  vector<int> mul(vector<int> a, vector<int> b) {
    int n = 1, m = a.size() + b.size() - 1;
    if (a.size() * b.size() <= 4096) {
      vector<int> c(m, 0);
      for (int i = 0; i < a.size(); i++)
        for (int j = 0; j < b.size(); j++)
          c[i + j] = (c[i + j] + 1ll * a[i] * b[j] % mod) % mod;
      return c;
    }
    while (n < m) n <<= 1;
    a.resize(n), b.resize(n);
    ntt(a, 0), ntt(b, 0);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * b[i] % mod;
    ntt(a, 1);
    a.resize(m);
    return a;
  }
  vector<int> inv(vector<int> a) {
    int n = a.size();
    if (n == 1) {
      a[0] = qp(a[0], mod - 2);
      return a;
    }
    int m = 1;
    while (m < (n << 1)) m <<= 1;
    vector<int> a0 = a;
    a0.resize((n + 1) / 2);
    vector<int> b0 = inv(a0);
    a.resize(m), b0.resize(m);
    ntt(a, 0), ntt(b0, 0);
    for (int i = 0; i < m; i++)
      a[i] = 1ll * b0[i] * (2 - 1ll * a[i] * b0[i] % mod + mod) % mod;
    ntt(a, 1);
    a.resize(n);
    return a;
  }
  void dfs(int u) {
    vis[u] = 1;
    t++;
    if (!fl) return;
    if (!r[u]) return;
    if (b[u] || vis[r[u]]) b[u] = 1, fl = 0;
    dfs(r[u]);
    if (b[r[u]]) b[u] = 1, fl = 0;
  }
  int main() {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      int c;
      scanf("%d", &c);
      for (int j = 1; j <= c; j++) scanf("%d", &a[j]);
      for (int j = 2; j <= c; j++) {
        if (l[a[j]] && l[a[j]] != a[j - 1]) b[a[j]] = 1;
        l[a[j]] = a[j - 1];
      }
      for (int j = 1; j < c; j++) {
        if (r[a[j]] && r[a[j]] != a[j + 1]) b[a[j]] = 1;
        r[a[j]] = a[j + 1];
      }
    }
    mx = m + 5;
    s.resize(mx);
    for (int i = 1; i <= k; i++) {
      if (vis[i] || b[i]) continue;
      if (!l[i] && !r[i]) {
        s[1]++;
        vis[i] = 1;
        continue;
      }
      if (l[i] == i && r[i] == i) {
        vis[i] = b[i] = 1;
        continue;
      }
      if (!l[i] && r[i]) {
        t = 0, fl = 1;
        dfs(i);
        if (fl && t <= m) s[t]++;
      }
    }
    vector<int> f = vector<int>(mx, 0);
    f[0] = (f[0] + mod - 1) % mod;
    s[0] = (s[0] + mod - 1) % mod;
    vector<int> iv = inv(s);
    iv[0] = (iv[0] + mod - 1) % mod;
    f = mul(f, iv);
    printf("%d\n", f[m]);
    return 0;
  }
  