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.
/*
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Don't let that studying and grades consume you
Just live life young
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1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
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I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
Shining through the city with a little funk and soul
*/
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1408E
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int M = Integer.parseInt(st.nextToken());
         int N = Integer.parseInt(st.nextToken());
         int[] arr = new int[M];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < M; i++)
            arr[i] = Integer.parseInt(st.nextToken());
         int[] brr = new int[N];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < N; i++)
            brr[i] = Integer.parseInt(st.nextToken());
         ArrayList<Edge> ls = new ArrayList<Edge>();
         for(int i=0; i < M; i++)
         {
            st = new StringTokenizer(infile.readLine());
            int X = Integer.parseInt(st.nextToken());
            while(X-->0)
            {
               int j = Integer.parseInt(st.nextToken())-1;
               ls.add(new Edge(i, j+M, arr[i]+brr[j]));
            }
         }
         Collections.sort(ls);
         DSU union = new DSU(N+M+1);
         long res = 0L;
         for(Edge e: ls)
         {
            if(union.find(e.from) != union.find(e.to))
               union.merge(e.from, e.to);
            else
               res += e.cost;
         }
         System.out.println(res);
      }
   }
   class DSU
   {
      public int[] dsu;
      
      public DSU(int N)
      {
         dsu = new int[N+1];
         for(int i=0; i <= N; i++)
            dsu[i] = i;
      }
      public int find(int x) 
      {
   		return dsu[x] == x ? x : (dsu[x] = find(dsu[x]));
   	}
      public void merge(int x, int y) 
      {
   		int fx = find(x);
   		int fy = find(y);
   		dsu[fx] = fy;
   	}
   }
   class Edge implements Comparable<Edge>
   {
      public int from;
      public int to;
      public int cost;
      
      public Edge(int a, int b, int c)
      {
         from = a;
         to = b;
         cost = c;
      }
      public int compareTo(Edge oth)
      {
         return oth.cost-cost;
      }
   }