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 INF = 1000000007;
const long long INFL = 1000000000000000007;
const long double INFS = 0.00000001;
const long long MOD = 1000000007;
const long double PI = 3.14159265;
const long long dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
const long long dir2[8][2] = {{1, 1},  {1, 0},  {1, -1}, {0, 1},
                              {0, -1}, {-1, 1}, {-1, 0}, {-1, -1}};
long long n, m;
pair<long long, long long> a[100001];
long long b[100001];
long long sz[100001];
vector<pair<long long, long long> > ary[100001];
vector<pair<long long, pair<long long, long long> > > lst;
long long app[100001];
long long ans = 0;
struct UNIONFIND {
  struct node {
    long long r, f;
  } opt[100001];
  void ini() {
    for (long long i = 1; i <= 100000; i++) opt[i].f = i, opt[i].r = 1;
  }
  long long fd(long long x) {
    if (x != opt[x].f) opt[x].f = fd(opt[x].f);
    return opt[x].f;
  }
  void mdf(long long x, long long y) {
    x = fd(x), y = fd(y);
    if (x == y) return;
    if (opt[x].r < opt[y].r)
      opt[x].f = y, opt[y].r += opt[x].r;
    else
      opt[y].f = x, opt[x].r += opt[y].r;
  }
} ds;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j;
  ds.ini();
  cin >> n >> m;
  for (i = 1; i <= n; i++) cin >> a[i].first, a[i].second = i;
  for (i = 1; i <= m; i++) cin >> b[i];
  for (i = 1; i <= n; i++) {
    cin >> sz[i];
    for (j = 1; j <= sz[i]; j++) {
      long long x;
      cin >> x;
      ary[i].push_back({b[x], x});
      lst.push_back({a[i].first + b[x], {i, x}});
    }
    sort(ary[i].begin(), ary[i].end());
  }
  sort(a + 1, a + n + 1);
  sort(lst.begin(), lst.end());
  for (i = lst.size() - 1; i >= 0; i--) {
    long long f = lst[i].second.first;
    long long f2 = lst[i].second.second;
    bool add = true;
    if (app[f2]) {
      if (ds.fd(app[f2]) != ds.fd(f))
        ds.mdf(app[f2], f);
      else {
        ans += lst[i].first;
        add = false;
      }
    }
    if (add) app[f2] = f;
  }
  cout << ans << endl;
  return 0;
}
