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 dv = 998244353;
  long long n, m, d, nn = 0, dsu[300069], cc[300069], ce[300069], dg[2][300069],
                     cdg[2][300069], fq[300069], dp[300069], ex[300069];
  map<long long, long long> ve[300069];
  long long fd(long long x) {
    if (dsu[x] != x) {
      dsu[x] = fd(dsu[x]);
    }
    return dsu[x];
  }
  int main() {
    long long i, j, ii, k, l, sz;
    scanf("%lld%lld%lld", &n, &m, &d);
    for (i = 1; i <= d; i++) {
      dsu[i] = i;
      cc[i] = 1;
    }
    for (i = 0; i < n; i++) {
      scanf("%lld", &sz);
      for (j = 0; j < sz; j++) {
        scanf("%lld", &k);
        if (j && !ve[l][k]) {
          ve[l][k] = 1;
          if (fd(k) != fd(l)) {
            cc[fd(l)] += cc[fd(k)];
            ce[fd(l)] += ce[fd(k)];
            for (ii = 0; ii < 2; ii++) {
              cdg[ii][fd(l)] += cdg[ii][fd(k)];
            }
            dsu[fd(k)] = l;
          }
          ce[fd(k)]++;
          for (ii = 0; ii < 2; ii++) {
            cdg[ii][fd(k)] -= dg[ii][k] == 1;
            dg[ii][k]++;
            cdg[ii][fd(k)] += dg[ii][k] == 1;
            swap(k, l);
          }
        }
        l = k;
      }
    }
    for (i = 1; i <= d; i++) {
      if (fd(i) == i && ce[i] == cc[i] - 1 && cdg[0][i] == cc[i] - 1 &&
          cdg[1][i] == cc[i] - 1) {
        fq[cc[i]]++;
      }
    }
    for (i = 1; i <= d; i++) {
      if (fq[i]) {
        nn++;
        ex[nn] = i;
      }
    }
    dp[0] = 1;
    for (i = 1; i <= m; i++) {
      for (j = 1; j <= nn; j++) {
        if (i >= ex[j]) {
          dp[i] = (dp[i] + dp[i - ex[j]] * fq[ex[j]]) % dv;
        }
      }
    }
    printf("%lld\n", dp[m]);
  }
  