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("O3", "unroll-all-loops")
#pragma GCC target("sse4.2")
using namespace std;
ifstream in;
ofstream out;
const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = ml * kk + 7;
const long long inf = ml * ml * ml + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long n, m, ans;
vector<long long> a, b;
set<pair<long long, pair<long long, long long>>> e;
vector<long long> p, sz;
bool viv = false;
long long pp(long long v) {
  if (v == p[v]) return v;
  return p[v] = pp(p[v]);
}
void unit(long long v, long long u, long long to_add) {
  v = pp(v);
  u = pp(u);
  if (v == u) {
    ans -= to_add;
    return;
  }
  if (sz[v] < sz[u]) swap(v, u);
  sz[v] += sz[u];
  p[u] = v;
}
void adder() {
  auto mini = *e.begin();
  e.erase(mini);
  unit(mini.second.first, mini.second.second, mini.first);
}
void solve() {
  cin >> m >> n;
  a.resize(m);
  b.resize(n);
  for (auto &i : a) cin >> i;
  for (auto &i : b) cin >> i;
  for (int i = 0; i < m; i++) {
    long long k;
    cin >> k;
    for (int j = 0; j < k; j++) {
      long long v = i;
      long long u;
      cin >> u;
      u--;
      e.insert({-(a[v] + b[u]), {v, u + m}});
    }
  }
  p.resize(n + m);
  sz.resize(n + m, 1);
  for (int i = 0; i < p.size(); i++) p[i] = i;
  while (e.size()) adder();
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
