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;
const int maxp = 22;
const long double EPS = 1e-18;
const long long INF = 1e18;
const int MOD = 1e9 + 7;
const int N = 2e5 + 1;
struct DSU {
  vector<int> par, siz;
  int cnt;
  void init(int n) {
    cnt = n;
    par.assign(n + 1, 0);
    siz.assign(n + 1, 1);
    iota(par.begin(), par.end(), 0);
  }
  int find(int x) { return (par[x] == x) ? x : par[x] = find(par[x]); }
  bool merge(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return false;
    if (siz[x] < siz[y]) swap(x, y);
    siz[x] += siz[y];
    par[y] = x;
    cnt--;
    return true;
  }
};
struct ano {
  int x, y, z;
  bool operator<(const ano &a) const { return z > a.z; }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m, n;
  cin >> m >> n;
  int a[m], b[n];
  int i;
  for (i = 0; i < m; i++) cin >> a[i];
  for (i = 0; i < n; i++) cin >> b[i];
  long long tot = 0;
  vector<ano> arr;
  for (i = 0; i < m; i++) {
    int s;
    cin >> s;
    for (int j = 0; j < s; j++) {
      int x;
      cin >> x;
      int cost = b[x - 1] + a[i];
      arr.push_back({x, i + n + 1, cost});
      tot += cost;
    }
  }
  sort((arr).begin(), (arr).end());
  long long mst = 0;
  DSU dsu;
  dsu.init(n + m);
  for (auto x : arr) {
    if (dsu.merge(x.x, x.y)) {
      mst += x.z;
    }
  }
  cout << tot - mst << '\n';
}
