#include <bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr LL mod = 998244353;
void no() {
  cout << "0";
  exit(0);
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  vector<set<int>> G(k + 1), H(k + 1);
  for (int i = 0; i < n; i += 1) {
    int c;
    cin >> c;
    vector<int> A(c);
    for (int& x : A) cin >> x;
    for (int j = 1; j < c; j += 1) {
      G[A[j - 1]].insert(A[j]);
      H[A[j]].insert(A[j - 1]);
    }
  }
  map<int, int> mp;
  for (int i = 1; i <= k; i += 1)
    if (H[i].empty()) {
      int x = 0;
      for (int j = i; j;) {
        x += 1;
        if (H[j].size() > 1) {
          x = -1;
          break;
        }
        if (G[j].empty()) break;
        if (G[j].size() > 1) {
          x = -1;
          break;
        }
        j = *G[j].begin();
      }
      if (x != -1) mp[x] += 1;
    }
  vector<LL> dp(m + 1);
  dp[0] = 1;
  for (int i = 1; i <= m; i += 1)
    for (auto [x, y] : mp)
      if (i >= x) dp[i] = (dp[i] + dp[i - x] * y) % mod;
  cout << dp[m];
  return 0;
}

### Prompt

Create a solution in cpp for 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;
using LL = long long;
constexpr LL mod = 998244353;
void no() {
  cout << "0";
  exit(0);
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  vector<set<int>> G(k + 1), H(k + 1);
  for (int i = 0; i < n; i += 1) {
    int c;
    cin >> c;
    vector<int> A(c);
    for (int& x : A) cin >> x;
    for (int j = 1; j < c; j += 1) {
      G[A[j - 1]].insert(A[j]);
      H[A[j]].insert(A[j - 1]);
    }
  }
  map<int, int> mp;
  for (int i = 1; i <= k; i += 1)
    if (H[i].empty()) {
      int x = 0;
      for (int j = i; j;) {
        x += 1;
        if (H[j].size() > 1) {
          x = -1;
          break;
        }
        if (G[j].empty()) break;
        if (G[j].size() > 1) {
          x = -1;
          break;
        }
        j = *G[j].begin();
      }
      if (x != -1) mp[x] += 1;
    }
  vector<LL> dp(m + 1);
  dp[0] = 1;
  for (int i = 1; i <= m; i += 1)
    for (auto [x, y] : mp)
      if (i >= x) dp[i] = (dp[i] + dp[i - x] * y) % mod;
  cout << dp[m];
  return 0;
}

```