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 long long SIZE = 1e6 * 2 + 10, INF = 1e9 * 1e9 + 10;
vector<vector<long long>> el, gr;
long long dsu[SIZE];
long long par(int v) {
  if (dsu[v] == v)
    return v;
  else
    return dsu[v] = par(dsu[v]);
}
void uni(int v, int u) {
  u = par(u);
  v = par(v);
  if (u != v) {
    dsu[u] = v;
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  ;
  long long n, m;
  cin >> m >> n;
  vector<long long> vec1(m), vec2(n);
  vector<pair<long long, long long>> srt;
  long long mx = 0;
  long long i = 0, s1 = 0, s2 = 0;
  for (auto &cur : vec1) {
    cin >> cur;
    srt.push_back({cur, i});
    i++;
  }
  for (auto &cur : vec2) {
    cin >> cur;
  }
  for (int i = 0; i < n; i++) dsu[i] = i;
  sort(srt.rbegin(), srt.rend());
  long long ans = 0;
  el.resize(m);
  for (int i = 0; i < m; i++) {
    long long sz;
    cin >> sz;
    el[i].resize(sz);
    for (int j = 0; j < sz; j++) {
      cin >> el[i][j];
      el[i][j]--;
    }
  }
  set<long long> us;
  gr.resize(m);
  long long dis = 0;
  for (int i = 0; i < m; i++) {
    long long mx = 0;
    for (auto cur : el[srt[i].second]) {
      ans += vec2[cur] + srt[i].first;
      if (us.find(cur) == us.end()) {
        dis += srt[i].first + vec2[cur];
        us.insert(cur);
        gr[srt[i].second].push_back(cur);
      }
    }
  }
  vector<tuple<long long, long long, long long>> edges;
  for (int i = 0; i < m; i++) {
    if (gr[i].size() == 0) {
      pair<long long, long long> mx = {-1, -1};
      for (auto cur : el[i]) mx = max(mx, {vec2[cur], cur});
      dis += vec1[i] + mx.first;
      gr[i].push_back(mx.second);
    }
  }
  for (int i = 0; i < m; i++) {
    set<long long> s;
    if (gr[i].size() == 0) continue;
    for (int j = 1; j < gr[i].size(); j++) {
      uni(gr[i][j], gr[i][j - 1]);
    }
    for (auto cur : gr[i]) s.insert(cur);
    for (auto cur : el[i]) {
      if (s.find(cur) == s.end()) {
        edges.push_back({vec1[i] + vec2[cur], cur, gr[i][0]});
      }
    }
  }
  sort(edges.rbegin(), edges.rend());
  for (auto cur : edges) {
    long long u = par(get<1>(cur)), v = par(get<2>(cur)), c = get<0>(cur);
    if (u != v) {
      uni(u, v);
      dis += c;
    }
  }
  cout << ans - dis;
  return 0;
}
