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;
  long long get() {
    long long x = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)) {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = x * 10 + c - '0';
      c = getchar();
    }
    return x * f;
  }
  const long long N = 3e5 + 5, P = 998244353;
  long long n, m, k, pre[N], suf[N], f[N], a[N], ban[N], cnt[N], w[N], v[N], tot;
  signed main() {
    n = get(), m = get(), k = get();
    for (long long t = 1; t <= n; t++) {
      long long len = get();
      for (long long i = 1; i <= len; i++) a[i] = get();
      long long flag = 1;
      for (long long i = 2; i <= len; i++) {
        if (pre[a[i]] && pre[a[i]] != a[i - 1]) {
          flag = 0;
          continue;
        }
        if (suf[a[i - 1]] && suf[a[i - 1]] != a[i]) {
          flag = 0;
          continue;
        }
        pre[a[i]] = a[i - 1], suf[a[i - 1]] = a[i];
      }
      if (!flag)
        for (long long i = 1; i <= len; i++) ban[a[i]] = 1;
    }
    for (long long i = 1; i <= k; i++) {
      if (pre[i]) continue;
      long long flag = 0;
      for (long long now = i; now; now = suf[now]) flag |= ban[now];
      if (!flag) {
        long long len = 0;
        for (long long now = i; now; now = suf[now]) ++len;
        cnt[len]++;
      }
    }
    for (long long i = 1; i <= k; i++)
      if (cnt[i]) ++tot, w[tot] = i, v[tot] = cnt[i];
    f[0] = 1;
    for (long long i = 0; i <= m; i++)
      for (long long j = 1; j <= tot; j++)
        if (w[j] <= i) f[i] = (f[i] + v[j] * f[i - w[j]] % P) % P;
    printf("%lld\n", f[m]);
    return 0;
  }
  