#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10, MOD = 998244353;
struct Graph {
  vector<pair<int, long long> > son[N];
  int in[N];
  void add(int x, int y, long long w) {
    son[x].push_back(make_pair(y, w));
    in[y]++;
  }
  void init(int MAXN) {
    for (int i = 0; i <= MAXN; i++) son[i].clear(), in[i] = 0;
  }
} gr;
int n, m, k;
set<int> sson[N];
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    int c;
    cin >> c;
    int lst;
    for (int j = 1; j <= c; j++) {
      int x;
      cin >> x;
      if (j > 1 && !sson[lst].count(x)) {
        gr.add(lst, x, 1);
        sson[lst].insert(x);
      }
      lst = x;
    }
  }
  vector<bool> vis(k + 1);
  vector<int> cnt(k + 1);
  function<void(int, int)> dfs = [&](int u, int dep) {
    if (vis[u]) return;
    vis[u] = 1;
    if (gr.in[u] > 1) return;
    if (gr.son[u].size() > 1) return;
    if (gr.son[u].size() == 0) {
      cnt[dep]++;
      return;
    }
    dfs(gr.son[u][0].first, dep + 1);
    return;
  };
  for (int i = 1; i <= k; i++) {
    if (gr.in[i] == 0) dfs(i, 1);
  }
  vector<int> v;
  for (int i = 1; i <= k; i++)
    if (cnt[i]) {
      v.push_back(i);
    }
  vector<long long> dp(m + 1);
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (auto j : v) {
      if (i >= j) dp[i] = (dp[i] + dp[i - j] * cnt[j]) % MOD;
    }
  }
  cout << dp[m];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int ttt = 1;
  for (int i = 1; i <= ttt; i++) solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10, MOD = 998244353;
struct Graph {
  vector<pair<int, long long> > son[N];
  int in[N];
  void add(int x, int y, long long w) {
    son[x].push_back(make_pair(y, w));
    in[y]++;
  }
  void init(int MAXN) {
    for (int i = 0; i <= MAXN; i++) son[i].clear(), in[i] = 0;
  }
} gr;
int n, m, k;
set<int> sson[N];
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    int c;
    cin >> c;
    int lst;
    for (int j = 1; j <= c; j++) {
      int x;
      cin >> x;
      if (j > 1 && !sson[lst].count(x)) {
        gr.add(lst, x, 1);
        sson[lst].insert(x);
      }
      lst = x;
    }
  }
  vector<bool> vis(k + 1);
  vector<int> cnt(k + 1);
  function<void(int, int)> dfs = [&](int u, int dep) {
    if (vis[u]) return;
    vis[u] = 1;
    if (gr.in[u] > 1) return;
    if (gr.son[u].size() > 1) return;
    if (gr.son[u].size() == 0) {
      cnt[dep]++;
      return;
    }
    dfs(gr.son[u][0].first, dep + 1);
    return;
  };
  for (int i = 1; i <= k; i++) {
    if (gr.in[i] == 0) dfs(i, 1);
  }
  vector<int> v;
  for (int i = 1; i <= k; i++)
    if (cnt[i]) {
      v.push_back(i);
    }
  vector<long long> dp(m + 1);
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (auto j : v) {
      if (i >= j) dp[i] = (dp[i] + dp[i - j] * cnt[j]) % MOD;
    }
  }
  cout << dp[m];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int ttt = 1;
  for (int i = 1; i <= ttt; i++) solve();
  return 0;
}

```