#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> c(n);
  vector<vector<int>> A(n);
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    A[i].resize(c[i]);
    for (int j = 0; j < c[i]; j++) {
      cin >> A[i][j];
      A[i][j]--;
    }
  }
  vector<vector<int>> E1(k), E2(k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < c[i] - 1; j++) {
      E1[A[i][j]].push_back(A[i][j + 1]);
      E2[A[i][j + 1]].push_back(A[i][j]);
    }
  }
  for (int i = 0; i < k; i++) {
    sort(E1[i].begin(), E1[i].end());
    E1[i].erase(unique(E1[i].begin(), E1[i].end()), E1[i].end());
    sort(E2[i].begin(), E2[i].end());
    E2[i].erase(unique(E2[i].begin(), E2[i].end()), E2[i].end());
  }
  vector<bool> used(k, false);
  vector<int> S;
  for (int i = 0; i < k; i++) {
    if (!used[i] && E2[i].empty()) {
      used[i] = true;
      int v = i;
      int cnt = 1;
      bool ok = true;
      while (true) {
        if (E1[v].empty()) {
          break;
        }
        if (E1[v].size() >= 2) {
          ok = false;
          break;
        }
        int w = E1[v][0];
        if (E2[w].size() >= 2) {
          ok = false;
          break;
        }
        if (used[w]) {
          ok = false;
          break;
        }
        used[w] = true;
        cnt++;
        v = w;
      }
      if (ok) {
        S.push_back(cnt);
      }
    }
  }
  int cnt = S.size();
  map<int, int> mp;
  for (int i = 0; i < cnt; i++) {
    mp[S[i]]++;
  }
  int cnt2 = 0;
  vector<int> a, b;
  for (auto P : mp) {
    a.push_back(P.first);
    b.push_back(P.second);
    cnt2++;
  }
  vector<long long> dp(m + 1, 0);
  dp[0] = 1;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < cnt2; j++) {
      if (i + a[j] <= m) {
        dp[i + a[j]] += dp[i] * b[j];
        dp[i + a[j]] %= MOD;
      }
    }
  }
  cout << dp[m] << endl;
}

### Prompt

Your challenge is to write 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 long long MOD = 998244353;
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> c(n);
  vector<vector<int>> A(n);
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    A[i].resize(c[i]);
    for (int j = 0; j < c[i]; j++) {
      cin >> A[i][j];
      A[i][j]--;
    }
  }
  vector<vector<int>> E1(k), E2(k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < c[i] - 1; j++) {
      E1[A[i][j]].push_back(A[i][j + 1]);
      E2[A[i][j + 1]].push_back(A[i][j]);
    }
  }
  for (int i = 0; i < k; i++) {
    sort(E1[i].begin(), E1[i].end());
    E1[i].erase(unique(E1[i].begin(), E1[i].end()), E1[i].end());
    sort(E2[i].begin(), E2[i].end());
    E2[i].erase(unique(E2[i].begin(), E2[i].end()), E2[i].end());
  }
  vector<bool> used(k, false);
  vector<int> S;
  for (int i = 0; i < k; i++) {
    if (!used[i] && E2[i].empty()) {
      used[i] = true;
      int v = i;
      int cnt = 1;
      bool ok = true;
      while (true) {
        if (E1[v].empty()) {
          break;
        }
        if (E1[v].size() >= 2) {
          ok = false;
          break;
        }
        int w = E1[v][0];
        if (E2[w].size() >= 2) {
          ok = false;
          break;
        }
        if (used[w]) {
          ok = false;
          break;
        }
        used[w] = true;
        cnt++;
        v = w;
      }
      if (ok) {
        S.push_back(cnt);
      }
    }
  }
  int cnt = S.size();
  map<int, int> mp;
  for (int i = 0; i < cnt; i++) {
    mp[S[i]]++;
  }
  int cnt2 = 0;
  vector<int> a, b;
  for (auto P : mp) {
    a.push_back(P.first);
    b.push_back(P.second);
    cnt2++;
  }
  vector<long long> dp(m + 1, 0);
  dp[0] = 1;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < cnt2; j++) {
      if (i + a[j] <= m) {
        dp[i + a[j]] += dp[i] * b[j];
        dp[i + a[j]] %= MOD;
      }
    }
  }
  cout << dp[m] << endl;
}

```