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;
using ll = long long;
using P = pair<int, int>;
struct UnionFind {
  vector<int> d;
  UnionFind(int n = 0) : d(n, -1) {}
  int find(int x) {
    if (d[x] < 0) return x;
    return d[x] = find(d[x]);
  }
  bool unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return false;
    if (d[x] > d[y]) swap(x, y);
    d[x] += d[y];
    d[y] = x;
    return true;
  }
  bool same(int x, int y) { return find(x) == find(y); }
  int size(int x) { return -d[find(x)]; }
};
int main() {
  int m, n;
  cin >> m >> n;
  vector<int> a(m), b(n);
  for (int i = 0; i < (m); i++) cin >> a[i];
  for (int i = 0; i < (n); i++) cin >> b[i];
  vector<vector<int>> e(m);
  for (int i = 0; i < (m); i++) {
    int s;
    cin >> s;
    e[i] = vector<int>(s);
    for (int j = 0; j < (s); j++) {
      cin >> e[i][j];
      e[i][j]--;
    }
  }
  vector<pair<int, P>> ppv;
  for (int i = 0; i < (m); i++)
    for (int j : e[i]) ppv.push_back(make_pair(a[i] + b[j], make_pair(i, j)));
  sort(ppv.rbegin(), ppv.rend());
  vector<int> pre(m, -1);
  UnionFind uf(n);
  ll tot = 0;
  for (auto pp : ppv) {
    int c = pp.first, i = pp.second.first, j = pp.second.second;
    if (pre[i] == -1 || !uf.same(pre[i], j)) {
      if (pre[i] != -1) uf.unite(pre[i], j);
      pre[i] = j;
    } else
      tot += c;
  }
  cout << tot << endl;
  return 0;
}
