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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
const char dir[] = {'R', 'L', 'D', 'U'};
ll add(ll a, ll b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
int mult(int a, int b) { return ((ll)a * b) % MOD; }
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
struct union_find {
  vector<int> par;
  vector<int> sz;
  union_find(int n) {
    par.resize(n);
    sz.assign(n, 1);
    iota(par.begin(), par.end(), 0);
  }
  int find(int u) {
    if (par[u] != u) {
      par[u] = find(par[u]);
    }
    return par[u];
  }
  bool unite(int u, int v) {
    u = find(u);
    v = find(v);
    if (u == v) return false;
    if (sz[u] > sz[v]) {
      swap(u, v);
    }
    par[u] = v;
    sz[v] += sz[u];
    return true;
  }
};
bool comp(pair<int, pii> a, pair<int, pii> b) { return a.first > b.first; }
int main() {
  setIO();
  int m, n;
  cin >> m >> n;
  vector<pair<int, pii>> edges;
  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];
  }
  ll totCost = 0;
  for (int i = 0; i < m; ++i) {
    int sz;
    cin >> sz;
    for (int j = 0; j < sz; ++j) {
      int nd;
      cin >> nd;
      --nd;
      totCost += a[i] + b[nd];
      edges.emplace_back(make_pair(a[i] + b[nd], make_pair(nd, i + n)));
    }
  }
  sort((edges).begin(), (edges).end(), comp);
  union_find uf(n + m);
  for (auto i : edges) {
    if (uf.unite(i.second.first, i.second.second)) {
      totCost -= i.first;
    }
  }
  cout << totCost << "\n";
}
