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;
long long p[200001];
long long vertex[100001];
long long wt = 0;
bool cmp(pair<long long, pair<long long, long long>> &a,
         pair<long long, pair<long long, long long>> &b) {
  return a.first < b.first;
}
long long Root(long long x) {
  while (x != p[x]) {
    p[x] = p[p[x]];
    x = p[x];
  }
  return x;
}
void Union(long long x, long long y, long long w) {
  long long r_x = Root(x);
  long long r_y = Root(y);
  if (r_x != r_y) {
    p[r_x] = r_y;
    wt += w;
  }
}
signed main() {
  for (long long i = 0; i <= 200000; i++) p[i] = i;
  long long m, n;
  cin >> m >> n;
  long long par = m + 1;
  for (long long i = 1; i <= n; i++) {
    vertex[i] = par;
    par++;
  }
  long long a[m], b[n];
  for (long long i = 0; i < m; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) cin >> b[i];
  vector<pair<long long, pair<long long, long long>>> e;
  for (long long i = 0; i < m; i++) {
    long long sz;
    cin >> sz;
    for (long long j = 0; j < sz; j++) {
      long long y;
      cin >> y;
      e.push_back({a[i] + b[y - 1], {i + 1, vertex[y]}});
    }
  }
  sort(e.begin(), e.end(), cmp);
  long long tc = 0;
  for (long long i = (long long)e.size() - 1; i >= 0; i--) {
    long long x = e[i].second.first, y = e[i].second.second, w = e[i].first;
    Union(x, y, w);
    tc += w;
  }
  cout << tc - wt;
}
