#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> nxt(k, -1), pre(k, -1);
  for (int i = 0; i < n; ++i) {
    int c;
    cin >> c;
    int x;
    cin >> x;
    --x;
    for (int j = 1; j < c; ++j) {
      int y;
      cin >> y;
      --y;
      if (nxt[x] != -1 && nxt[x] != y)
        nxt[x] = -2;
      else
        nxt[x] = y;
      if (pre[y] != -1 && pre[y] != x)
        pre[y] = -2;
      else
        pre[y] = x;
      x = y;
    }
  }
  vector<bool> seen(k, 0);
  vector<int> counts(k + 1, 0);
  for (int i = 0; i < k; ++i) {
    if (pre[i] == -1) {
      int len = 0;
      bool works = 1;
      for (int x = i; x >= 0 && works; x = nxt[x], ++len) {
        if (pre[x] == -2 || nxt[x] == -2 || seen[x]) works = 0;
        seen[x] = 1;
      }
      if (works) ++counts[len];
    }
  }
  vector<pair<int, int>> prs;
  for (int i = 1; i <= k; ++i) {
    if (counts[i] >= 1) prs.emplace_back(i, counts[i]);
  }
  vector<ll> dp(m + 1, 0);
  dp[0] = 1;
  for (int i = 1; i <= m; ++i) {
    for (auto pr : prs) {
      if (pr.first <= i) dp[i] = (dp[i] + dp[i - pr.first] * pr.second) % MOD;
    }
  }
  cout << dp[m] << '\n';
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
const ll MOD = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> nxt(k, -1), pre(k, -1);
  for (int i = 0; i < n; ++i) {
    int c;
    cin >> c;
    int x;
    cin >> x;
    --x;
    for (int j = 1; j < c; ++j) {
      int y;
      cin >> y;
      --y;
      if (nxt[x] != -1 && nxt[x] != y)
        nxt[x] = -2;
      else
        nxt[x] = y;
      if (pre[y] != -1 && pre[y] != x)
        pre[y] = -2;
      else
        pre[y] = x;
      x = y;
    }
  }
  vector<bool> seen(k, 0);
  vector<int> counts(k + 1, 0);
  for (int i = 0; i < k; ++i) {
    if (pre[i] == -1) {
      int len = 0;
      bool works = 1;
      for (int x = i; x >= 0 && works; x = nxt[x], ++len) {
        if (pre[x] == -2 || nxt[x] == -2 || seen[x]) works = 0;
        seen[x] = 1;
      }
      if (works) ++counts[len];
    }
  }
  vector<pair<int, int>> prs;
  for (int i = 1; i <= k; ++i) {
    if (counts[i] >= 1) prs.emplace_back(i, counts[i]);
  }
  vector<ll> dp(m + 1, 0);
  dp[0] = 1;
  for (int i = 1; i <= m; ++i) {
    for (auto pr : prs) {
      if (pr.first <= i) dp[i] = (dp[i] + dp[i - pr.first] * pr.second) % MOD;
    }
  }
  cout << dp[m] << '\n';
}

```