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;
int read() {
  int x = 0;
  bool flg = false;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flg = true;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
  return flg ? -x : x;
}
int m, n;
int a[100010], b[100010];
vector<int> d[100010];
vector<int> e[100010];
bool cmp1(int x, int y) { return b[x] < b[y] || b[x] == b[y] && x < y; }
bool cmp2(int x, int y) { return a[x] < a[y] || a[x] == a[y] && x < y; }
int f[100010];
int find(int x) { return f[x] ? f[x] = find(f[x]) : x; }
struct Rec {
  int x, y;
} rec[200010];
int top;
bool operator<(Rec x, Rec y) { return a[x.x] + b[x.y] > a[y.x] + b[y.y]; }
int main() {
  m = read(), n = read();
  for (int i = 1; i <= m; i++) a[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  for (int i = 1; i <= m; i++) {
    int t = read();
    while (t--) {
      int x = read();
      d[i].push_back(x);
      e[x].push_back(i);
      rec[++top] = (Rec){i, x};
    }
  }
  for (int i = 1; i <= m; i++) sort(d[i].begin(), d[i].end(), cmp1);
  for (int i = 1; i <= n; i++) sort(e[i].begin(), e[i].end(), cmp2);
  long long ans = 0;
  sort(rec + 1, rec + top + 1);
  for (int i = 1; i <= top; i++) {
    int x = rec[i].x, y = rec[i].y;
    int t = lower_bound(e[y].begin(), e[y].end(), x, cmp2) - e[y].begin();
    if (t + 1 == e[y].size()) continue;
    int u = find(e[y][t]), v = find(e[y][t + 1]);
    if (u ^ v)
      f[u] = v;
    else
      ans += a[x] + b[y];
  }
  cout << ans << endl;
  return 0;
}
