You are given m sets of integers A_1, A_2, …, A_m; elements of these sets are integers between 1 and n, inclusive.

There are two arrays of positive integers a_1, a_2, …, a_m and b_1, b_2, …, b_n. 

In one operation you can delete an element j from the set A_i and pay a_i + b_j coins for that.

You can make several (maybe none) operations (some sets can become empty).

After that, you will make an edge-colored undirected graph consisting of n vertices. For each set A_i you will add an edge (x, y) with color i for all x, y ∈ A_i and x < y. Some pairs of vertices can be connected with more than one edge, but such edges have different colors.

You call a cycle i_1 → e_1 → i_2 → e_2 → … → i_k → e_k → i_1 (e_j is some edge connecting vertices i_j and i_{j+1} in this graph) rainbow if all edges on it have different colors.

Find the minimum number of coins you should pay to get a graph without rainbow cycles.

Input

The first line contains two integers m and n (1 ≤ m, n ≤ 10^5), the number of sets and the number of vertices in the graph.

The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ 10^9).

The third line contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ 10^9).

In the each of the next of m lines there are descriptions of sets. In the i-th line the first integer s_i (1 ≤ s_i ≤ n) is equal to the size of A_i. Then s_i integers follow: the elements of the set A_i. These integers are from 1 to n and distinct.

It is guaranteed that the sum of s_i for all 1 ≤ i ≤ m does not exceed 2 ⋅ 10^5.

Output

Print one integer: the minimum number of coins you should pay for operations to avoid rainbow cycles in the obtained graph.

Examples

Input


3 2
1 2 3
4 5
2 1 2
2 1 2
2 1 2


Output


11


Input


7 8
3 6 7 9 10 7 239
8 1 9 7 10 2 6 239
3 2 1 3
2 4 1
3 1 3 7
2 4 3
5 3 4 5 6 7
2 5 7
1 8


Output


66

Note

In the first test, you can make such operations:

  * Delete element 1 from set 1. You should pay a_1 + b_1 = 5 coins for that. 
  * Delete element 1 from set 2. You should pay a_2 + b_1 = 6 coins for that. 



You pay 11 coins in total. After these operations, the first and the second sets will be equal to \{2\} and the third set will be equal to \{1, 2\}.

So, the graph will consist of one edge (1, 2) of color 3.

In the second test, you can make such operations:

  * Delete element 1 from set 1. You should pay a_1 + b_1 = 11 coins for that. 
  * Delete element 4 from set 2. You should pay a_2 + b_4 = 13 coins for that. 
  * Delete element 7 from set 3. You should pay a_3 + b_7 = 13 coins for that. 
  * Delete element 4 from set 4. You should pay a_4 + b_4 = 16 coins for that. 
  * Delete element 7 from set 6. You should pay a_6 + b_7 = 13 coins for that. 



You pay 66 coins in total.

After these operations, the sets will be:

  * \{2, 3\}; 
  * \{1\}; 
  * \{1, 3\}; 
  * \{3\}; 
  * \{3, 4, 5, 6, 7\}; 
  * \{5\}; 
  * \{8\}. 



We will get the graph:

<image>

There are no rainbow cycles in it.
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > graf[2 * 100 * 1000 + 5];
pair<int, int> maxi[2 * 100 * 1000 + 5];
int rep[2 * 100 * 1000 + 5], roz[2 * 100 * 1000 + 5];
int grup[100 * 1000 + 5], tab[100 * 1000 + 5];
int szuk(int dom) {
  if (dom == rep[dom]) {
    return dom;
  }
  return rep[dom] = szuk(rep[dom]);
}
void unia(int a, int b) {
  if (roz[a] > roz[b]) {
    roz[a] += roz[b];
    rep[b] = a;
  } else {
    roz[b] += roz[a];
    rep[a] = b;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long suma = 0;
  int m, n;
  bool czy = 1;
  cin >> m >> n;
  for (int i = 1; i < m + 1; ++i) {
    cin >> grup[i];
  }
  for (int i = 1; i < n + 1; ++i) {
    cin >> tab[i];
  }
  for (int i = 1; i < n + m + 1; ++i) {
    rep[i] = i;
    roz[i] = 1;
    maxi[i] = {-2 * 1000 * 1000 * 1000 + 5, -2 * 1000 * 1000 * 1000 + 5};
  }
  for (int i = 1, s; i < m + 1; ++i) {
    cin >> s;
    for (int j = 1, a; j < s + 1; ++j) {
      cin >> a;
      graf[a].push_back({i + n, grup[i] + tab[a]});
      graf[i + n].push_back({a, grup[i] + tab[a]});
      suma += grup[i] + tab[a];
    }
  }
  while (czy == 1) {
    for (int i = 1, szef; i < n + m + 1; ++i) {
      szef = szuk(i);
      for (int j = 0; j < graf[i].size(); ++j) {
        if (szef != szuk(graf[i][j].first) &&
            maxi[szef].second < graf[i][j].second) {
          maxi[szef] = graf[i][j];
        }
      }
    }
    czy = 0;
    for (int i = 1, szef_a, szef_b; i < n + m + 1; ++i) {
      if (maxi[i].first != -2 * 1000 * 1000 * 1000 + 5) {
        szef_a = szuk(i);
        szef_b = szuk(maxi[i].first);
        if (szef_a != szef_b) {
          suma -= maxi[i].second;
          unia(szef_a, szef_b);
          czy = 1;
        }
      }
      maxi[i] = {-2 * 1000 * 1000 * 1000 + 5, -2 * 1000 * 1000 * 1000 + 5};
    }
  }
  cout << suma << "\n";
  return 0;
}
