#include <bits/stdc++.h>
using namespace std;
const int N = 2000100;
const long long mod = 998244353;
const long long inf = 2000000000000000000;
vector<int> a[N];
bool ban[N];
set<int> g[N];
set<int> in[N];
set<int> out[N];
bool used[N];
vector<int> order;
void dfs(int v) {
  order.push_back(v);
  used[v] = true;
  for (int to : g[v]) {
    if (!used[to]) dfs(to);
  }
}
long long dp[N];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    int len;
    cin >> len;
    a[i].resize(len, 0);
    set<int> st;
    for (int &j : a[i]) cin >> j, st.insert(j);
    if (st.size() < len) {
      for (int x : st) ban[x] = true;
    } else {
      for (int j = 1; j < a[i].size(); j++) {
        g[a[i][j - 1]].insert(a[i][j]);
        g[a[i][j]].insert(a[i][j - 1]);
        out[a[i][j - 1]].insert(a[i][j]);
        in[a[i][j]].insert(a[i][j - 1]);
      }
    }
  }
  map<int, int> block;
  for (int i = 1; i <= k; i++) {
    if (used[i]) continue;
    dfs(i);
    bool OK = true;
    if (order.size() > 1) {
      int s = 0, f = 0, mid = 0;
      for (int j : order) {
        if (ban[j]) OK = false;
        if (in[j].size() == 1 && out[j].size() == 1)
          mid++;
        else if (in[j].size() == 1 && out[j].size() == 0)
          f++;
        else if (in[j].size() == 0 && out[j].size() == 1)
          s++;
      }
      if (s != 1 || f != 1 || mid + 2 != order.size()) OK = false;
    } else
      OK = (!ban[i]);
    if (OK) block[order.size()]++;
    order.clear();
  }
  vector<pair<int, int> > v;
  for (auto cur : block) v.push_back(cur);
  dp[0] = 1ll;
  for (int i = 1; i <= m; i++) {
    dp[i] = 0ll;
    for (auto cur : v) {
      if (i >= cur.first) {
        dp[i] += dp[i - cur.first] * cur.second;
        dp[i] %= mod;
      }
    }
  }
  cout << dp[m] << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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 = 2000100;
const long long mod = 998244353;
const long long inf = 2000000000000000000;
vector<int> a[N];
bool ban[N];
set<int> g[N];
set<int> in[N];
set<int> out[N];
bool used[N];
vector<int> order;
void dfs(int v) {
  order.push_back(v);
  used[v] = true;
  for (int to : g[v]) {
    if (!used[to]) dfs(to);
  }
}
long long dp[N];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    int len;
    cin >> len;
    a[i].resize(len, 0);
    set<int> st;
    for (int &j : a[i]) cin >> j, st.insert(j);
    if (st.size() < len) {
      for (int x : st) ban[x] = true;
    } else {
      for (int j = 1; j < a[i].size(); j++) {
        g[a[i][j - 1]].insert(a[i][j]);
        g[a[i][j]].insert(a[i][j - 1]);
        out[a[i][j - 1]].insert(a[i][j]);
        in[a[i][j]].insert(a[i][j - 1]);
      }
    }
  }
  map<int, int> block;
  for (int i = 1; i <= k; i++) {
    if (used[i]) continue;
    dfs(i);
    bool OK = true;
    if (order.size() > 1) {
      int s = 0, f = 0, mid = 0;
      for (int j : order) {
        if (ban[j]) OK = false;
        if (in[j].size() == 1 && out[j].size() == 1)
          mid++;
        else if (in[j].size() == 1 && out[j].size() == 0)
          f++;
        else if (in[j].size() == 0 && out[j].size() == 1)
          s++;
      }
      if (s != 1 || f != 1 || mid + 2 != order.size()) OK = false;
    } else
      OK = (!ban[i]);
    if (OK) block[order.size()]++;
    order.clear();
  }
  vector<pair<int, int> > v;
  for (auto cur : block) v.push_back(cur);
  dp[0] = 1ll;
  for (int i = 1; i <= m; i++) {
    dp[i] = 0ll;
    for (auto cur : v) {
      if (i >= cur.first) {
        dp[i] += dp[i - cur.first] * cur.second;
        dp[i] %= mod;
      }
    }
  }
  cout << dp[m] << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}

```