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")
#pragma GCC target("sse4")
using namespace std;
const double PI = acos(-1.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long find_parent(long long parent[], long long a) {
  if (parent[a] >= 0)
    return parent[a] = find_parent(parent, parent[a]);
  else
    return a;
}
void merge_sets(long long parent[], long long a, long long b) {
  long long p1 = find_parent(parent, a), p2 = find_parent(parent, b);
  if (p1 == p2) return;
  if (abs(parent[p1]) < abs(parent[p2])) swap(p1, p2);
  parent[p1] -= abs(parent[p2]);
  parent[p2] = p1;
}
const long long N = 1e5 + 10;
long long a[N], b[N];
long long n, m;
vector<pair<long long, long long>> v;
vector<long long> cur[N];
long long parent[N];
bool cmp(pair<long long, long long> p, pair<long long, long long> q) {
  long long cst1 = a[p.first] + b[p.second];
  long long cst2 = a[q.first] + b[q.second];
  return cst1 > cst2;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> m >> n;
  memset(parent, -1, sizeof(parent));
  ;
  for (long long i = 1; i <= m; i++) {
    cin >> a[i];
  }
  for (long long j = 1; j <= n; j++) {
    cin >> b[j];
  }
  long long tot = 0;
  for (long long i = 1; i <= m; i++) {
    long long s;
    cin >> s;
    for (long long j = 0; j < s; j++) {
      long long x;
      cin >> x;
      v.push_back({i, x});
      tot += a[i] + b[x];
    }
  }
  sort(v.begin(), v.end(), cmp);
  for (auto x : v) {
    long long ind = x.second;
    if (cur[x.first].empty() ||
        find_parent(parent, cur[x.first].back()) != find_parent(parent, ind)) {
      if (!cur[x.first].empty()) {
        merge_sets(parent, cur[x.first].back(), ind);
      }
      cur[x.first].push_back(x.second);
      tot -= (a[x.first] + b[x.second]);
    }
  }
  cout << tot << "\n";
}
