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 M = 300005;
  const int MOD = 998244353;
  int read() {
    int x = 0, f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9') {
      if (c == '-') f = -1;
    }
    while (c >= '0' && c <= '9') {
      x = (x << 3) + (x << 1) + (c ^ 48);
      c = getchar();
    }
    return x * f;
  }
  int n, m, k, t, tot, f[M], a[M], in[M], out[M], cnt[M], dp[M];
  int pd, zxy, sz, vis[M];
  map<pair<int, int>, int> mp;
  struct edge {
    int v, next;
  } e[2 * M];
  void add(int x, int y) {
    if (mp[make_pair(x, y)]) return;
    mp[make_pair(x, y)] = 1;
    in[y]++;
    out[x]++;
    e[++tot] = edge{y, f[x]}, f[x] = tot;
    e[++tot] = edge{x, f[y]}, f[y] = tot;
  }
  void dfs(int u) {
    sz++;
    vis[u] = 1;
    if (!in[u]) zxy = 1;
    if (in[u] >= 2 || out[u] >= 2) pd = 0;
    for (int i = f[u]; i; i = e[i].next) {
      int v = e[i].v;
      if (!vis[v]) dfs(v);
    }
  }
  signed main() {
    n = read();
    m = read();
    k = read();
    for (int i = 1; i <= n; i++) {
      int len = read();
      for (int j = 1, ls; j <= len; j++) {
        int x = read();
        if (j > 1) add(ls, x);
        ls = x;
      }
    }
    for (int i = 1; i <= k; i++)
      if (!vis[i]) {
        pd = 1;
        zxy = sz = 0;
        dfs(i);
        if (!pd || !zxy) continue;
        a[++t] = sz;
        cnt[sz]++;
      }
    sort(a + 1, a + 1 + t);
    t = unique(a + 1, a + 1 + t) - a - 1;
    dp[0] = 1;
    for (int i = 1; i <= m; i++)
      for (int j = 1; j <= t; j++)
        if (a[j] <= i) dp[i] = (dp[i] + 1ll * cnt[a[j]] * dp[i - a[j]]) % MOD;
    printf("%d\n", dp[m]);
  }
  