#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  set<int> v[k + 1];
  vector<int> in(k + 1, 0);
  vector<int> out(k + 1, 0);
  auto add_edge = [&](int a, int b) {
    if (v[a].find(b) != v[a].end()) return;
    v[a].insert(b);
    in[b]++;
    out[a]++;
  };
  for (int i = 1; i <= n; i++) {
    int c;
    cin >> c;
    int s;
    cin >> s;
    for (int j = 2; j <= c; j++) {
      int a;
      cin >> a;
      add_edge(s, a);
      s = a;
    }
  }
  vector<bool> vis(k + 1, 0);
  vector<int> cnt(m + 1, 0);
  for (int i = 1; i <= k; i++) {
    if (vis[i] || in[i] >= 1 || out[i] > 1) continue;
    bool ok = 1;
    int a = i;
    int len = 0;
    while (1) {
      ok &= (in[a] <= 1 && out[a] <= 1 && !vis[a]);
      if (!ok) break;
      vis[a] = 1;
      len++;
      if (out[a] == 0) break;
      a = (*v[a].begin());
    }
    if (ok && len <= m) cnt[len]++;
  }
  vector<array<int, 2>> opt;
  for (int i = 1; i <= m; i++)
    if (cnt[i] > 0) opt.push_back({i, cnt[i]});
  const long long mod = 998244353;
  auto add = [&](long long a, long long b) -> long long {
    return (a + b) % mod;
  };
  auto mul = [&](long long a, long long b) -> long long {
    return (a * b) % mod;
  };
  vector<long long> dp(m + 1, 0);
  dp[0] = 1;
  for (int i = 0; i <= m; i++) {
    for (auto [c, o] : opt)
      if (i + c <= m) dp[i + c] = add(dp[i + c], mul(o, dp[i]));
  }
  cout << dp[m] << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  set<int> v[k + 1];
  vector<int> in(k + 1, 0);
  vector<int> out(k + 1, 0);
  auto add_edge = [&](int a, int b) {
    if (v[a].find(b) != v[a].end()) return;
    v[a].insert(b);
    in[b]++;
    out[a]++;
  };
  for (int i = 1; i <= n; i++) {
    int c;
    cin >> c;
    int s;
    cin >> s;
    for (int j = 2; j <= c; j++) {
      int a;
      cin >> a;
      add_edge(s, a);
      s = a;
    }
  }
  vector<bool> vis(k + 1, 0);
  vector<int> cnt(m + 1, 0);
  for (int i = 1; i <= k; i++) {
    if (vis[i] || in[i] >= 1 || out[i] > 1) continue;
    bool ok = 1;
    int a = i;
    int len = 0;
    while (1) {
      ok &= (in[a] <= 1 && out[a] <= 1 && !vis[a]);
      if (!ok) break;
      vis[a] = 1;
      len++;
      if (out[a] == 0) break;
      a = (*v[a].begin());
    }
    if (ok && len <= m) cnt[len]++;
  }
  vector<array<int, 2>> opt;
  for (int i = 1; i <= m; i++)
    if (cnt[i] > 0) opt.push_back({i, cnt[i]});
  const long long mod = 998244353;
  auto add = [&](long long a, long long b) -> long long {
    return (a + b) % mod;
  };
  auto mul = [&](long long a, long long b) -> long long {
    return (a * b) % mod;
  };
  vector<long long> dp(m + 1, 0);
  dp[0] = 1;
  for (int i = 0; i <= m; i++) {
    for (auto [c, o] : opt)
      if (i + c <= m) dp[i + c] = add(dp[i + c], mul(o, dp[i]));
  }
  cout << dp[m] << "\n";
  return 0;
}

```