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 int mod = 998244353;
  struct DSU {
    int n;
    vector<int> link, sz;
    DSU(int n) : n(n), link(n), sz(n) {
      for (int i = 0; i < n; i++) {
        link[i] = i;
        sz[i] = 1;
      }
    }
    int first(int x) { return (x == link[x] ? x : link[x] = first(link[x])); }
    void unite(int x, int y) {
      x = first(x), y = first(y);
      if (x == y) {
        return;
      }
      if (sz[x] < sz[y]) {
        swap(x, y);
      }
      sz[x] += sz[y], link[y] = x;
    }
  };
  int n, m, k, dp[300001], nxt[300000], prv[300000];
  vector<int> arr[300000], dsc[300000];
  vector<pair<int, int>> trns;
  map<int, int> chns;
  bool bad[300000], appeared[300000];
  int mul(int x, int y) { return ((long long)x * y) % mod; }
  int main() {
    ios_base::sync_with_stdio(false);
    cin >> n >> m >> k;
    DSU grp(k);
    memset(nxt, -1, sizeof(nxt));
    memset(prv, -1, sizeof(prv));
    for (int i = 0, sz; i < n; i++) {
      cin >> sz;
      arr[i].resize(sz);
      for (int j = 0; j < sz; j++) {
        cin >> arr[i][j];
        appeared[arr[i][j] - 1] = true;
      }
      for (int j = 0; j < sz - 1; j++) {
        grp.unite(arr[i][j] - 1, arr[i][j + 1] - 1);
      }
    }
    for (int i = 0; i < k; i++) {
      dsc[grp.first(i)].push_back(i);
    }
    for (int i = 0; i < n; i++) {
      int sz = arr[i].size();
      for (int j = 0; j < sz - 1; j++) {
        if (prv[arr[i][j + 1] - 1] != -1 &&
            prv[arr[i][j + 1] - 1] != arr[i][j] - 1) {
          bad[grp.first(arr[i][j] - 1)] = true;
        }
        if (nxt[arr[i][j] - 1] != -1 && nxt[arr[i][j] - 1] != arr[i][j + 1] - 1) {
          bad[grp.first(arr[i][j] - 1)] = true;
        }
        prv[arr[i][j + 1] - 1] = arr[i][j] - 1;
        nxt[arr[i][j] - 1] = arr[i][j + 1] - 1;
      }
    }
    for (int i = 0; i < k; i++) {
      if (bad[i]) {
        for (int j : dsc[i]) {
          bad[j] = true;
        }
      }
    }
    for (int i = 0; i < k; i++) {
      if (prv[i] == -1 && !bad[i]) {
        int chn = 1, cur = i;
        while (nxt[cur] != -1) {
          cur = nxt[cur];
          ++chn;
        }
        ++chns[chn];
      }
    }
    for (auto it : chns) {
      trns.push_back(it);
    }
    dp[0] = 1;
    for (int i = 1; i <= m; i++) {
      for (auto j : trns) {
        if (i >= j.first) {
          dp[i] = (dp[i] + mul(dp[i - j.first], j.second)) % mod;
        }
      }
    }
    cout << dp[m] << '\n';
  }
  