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 maxn = 200100;
int N, M, A[maxn], B[maxn];
int par[maxn], sz[maxn];
int anc(int u) { return par[u] == u ? u : par[u] = anc(par[u]); }
bool merge(int u, int v) {
  u = anc(u);
  v = anc(v);
  if (u == v) return 0;
  if (sz[u] < sz[v]) swap(u, v);
  sz[u] += sz[v];
  par[v] = u;
  return 1;
}
struct edge {
  int u, v, w;
  edge() {}
  edge(int u, int v, int w) : u(u), v(v), w(w) {}
  bool operator<(const edge &other) const { return w > other.w; }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> M >> N;
  for (int i = 1; i <= M; ++i) cin >> A[i];
  for (int i = 1; i <= N; ++i) cin >> B[i];
  for (int i = 1; i <= M + N; ++i) {
    par[i] = i;
    sz[i] = 1;
  }
  vector<edge> Ed;
  using ll = long long;
  ll tot = 0;
  for (int i = 1; i <= M; ++i) {
    int S;
    cin >> S;
    for (int X, j = 1; j <= S; ++j) {
      cin >> X;
      tot += A[i] + B[X];
      Ed.emplace_back(i, X + M, A[i] + B[X]);
    }
  }
  sort(Ed.begin(), Ed.end());
  for (edge E : Ed) {
    if (merge(E.u, E.v)) tot -= E.w;
  }
  cout << tot << '\n';
  return 0;
}
