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("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("Ofast")
using namespace std;
const int INF = 1e9 + 10;
const long long INFll = 2e18;
const int BASE1 = 179;
const int BASE2 = 653;
const long long MOD = 998244353;
const int MAXA = 2000001;
const int MAXN = 2e5 + 1;
const int MAXLOG = 22;
const long long COST = 1000 * 1000 * 1000;
const long double PI = 3.1415926535;
const long double EPS = 1e-15;
struct edge {
  long long u, v, cost;
  edge(long long a, long long b, long long c) {
    u = a;
    v = b;
    cost = c;
  }
};
bool cmp(edge& a, edge& b) { return a.cost < b.cost; }
struct dsu {
  vector<int> p, d;
  dsu(int n) {
    p.resize(n);
    d.resize(n);
    for (int i = 0; i < (n); i++) p[i] = i;
  }
  int getp(int v) {
    if (p[v] == v) return v;
    return p[v] = getp(p[v]);
  }
  void unite(int u, int v) {
    u = getp(u), v = getp(v);
    if (d[u] < d[v]) swap(u, v);
    p[v] = u;
    if (d[u] == d[v]) d[u]++;
  }
};
void solve() {
  long long m, n;
  cin >> m >> n;
  vector<long long> a(m), b(n);
  for (int i = 0; i < (m); i++) cin >> a[i];
  for (int j = 0; j < (n); j++) cin >> b[j];
  vector<edge> edges;
  long long allcost = 0;
  for (int v = 0; v < (m); v++) {
    long long s;
    cin >> s;
    for (int i = 0; i < (s); i++) {
      long long u;
      cin >> u;
      u--;
      allcost += a[v] + b[u];
      edges.emplace_back(edge(u, v + n, a[v] + b[u]));
    }
  }
  sort((edges).rbegin(), (edges).rend(), cmp);
  dsu d(n + m);
  long long res = 0;
  for (edge e : edges) {
    if (d.getp(e.u) != d.getp(e.v)) {
      res += e.cost;
      d.unite(e.u, e.v);
    }
  }
  cout << allcost - res << "\n";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}
