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.
// Added -> dsu, mp, ncr, pair, sieve, suffix, zalgo
// To add -> Geometry, Advance DS Topics, Articulation Points & Bridges, Tries, LCA, Segment Tree

import java.io.*;
import java.util.*;
 
public class utkarsh {

    public static void main(String[] args) {
        new utkarsh().run();
    }
    
    BufferedReader br;
    PrintWriter out;
    
    long mod = (long) (1e9 + 7), inf = (long) (3e18);

    class edge {
        int U, V, W;
        edge(int u, int v, int w) {
            U = u;  V = v;  W = w;
        }
    }

    int parent[];
    
    void makeSet(int n) {
        parent = new int[n];
        for(int i = 0; i < n; i++)  parent[i] = i;
    }
    
    int findSet(int u) {
        if(u == parent[u])  return u;
        parent[u] = findSet(parent[u]);
        return parent[u];
    }
    
    boolean unionSet(int u, int v) {
        int x = findSet(u);
        int y = findSet(v);
        if(x == y)  return false;
        parent[x] = y;
        return true;
    }

    void solve() {
        int n = ni(), m = ni();
        int a[] = new int[n];
        for(int i = 0; i < n; i++)  a[i] = ni();
        int b[] = new int[m];
        for(int i = 0; i < m; i++)  b[i] = ni();
        ArrayList <edge> edges = new ArrayList<>();
        for(int i = 0; i < n; i++) {
            int size = ni();
            for(int j = 0; j < size; j++) {
                int x = ni() - 1;
                int u = i, v = n + x, w = a[i] + b[x];
                edges.add(new edge(u, v, w));
            }
        }
        Collections.sort(edges, (edge x, edge y) -> (y.W - x.W));
        makeSet(n+m);
        long ans = 0;
        for(edge e : edges) {
            if(! unionSet(e.U, e.V))    ans += e.W;
        }
        out.println(ans);
    }
    
    // -------- I/O Template -------------
    
    char nc() {
        return ns().charAt(0);
    }
    
    String nLine() {
        try {
            return br.readLine();
        } catch(IOException e) {
            return "-1";
        }
    }
    
    double nd() {
        return Double.parseDouble(ns());
    }
    
    long nl() {
        return Long.parseLong(ns());
    }
    
    int ni() {
        return Integer.parseInt(ns());
    }

    int[] na(int n) {
        int a[] = new int[n];
        for(int i = 0; i < n; i++)  a[i] = ni();
        return a;
    }
    
    StringTokenizer ip;
    
    String ns() {
        if(ip == null || !ip.hasMoreTokens()) {
            try {
                ip = new StringTokenizer(br.readLine());
            } catch(IOException e) {
                throw new InputMismatchException();
            }
        }
        return ip.nextToken();
    }

    void run() {
        try {
            if (System.getProperty("ONLINE_JUDGE") == null) {
                br = new BufferedReader(new FileReader("input.txt"));
                out = new PrintWriter("output.txt");
            } else {
                br = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
            }
        } catch (FileNotFoundException e) {
            System.out.println(e);
        }
        solve();
        out.flush();
    }
}