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

	public static void main(String[] args) throws Exception{
		int num = 998244353;

		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
 		StringTokenizer st = new StringTokenizer(bf.readLine());
 		int m = Integer.parseInt(st.nextToken());
 		int n = Integer.parseInt(st.nextToken());
 		long[] costs = new long[m+n+1];
 		StringTokenizer sta = new StringTokenizer(bf.readLine());
 		for(int j= 0;j<m;j++){
 			costs[j+n+1] = Long.parseLong(sta.nextToken());
 		}
 		StringTokenizer stb = new StringTokenizer(bf.readLine());
 		for(int j =0;j<n;j++){
 			costs[j+1] = Long.parseLong(stb.nextToken());
 		}
 		Map<Integer, ArrayList<Integer>> graph = new HashMap<Integer, ArrayList<Integer>>();
 		ArrayList<Edge> edges = new ArrayList<Edge>();
 		long sum = 0;
 		for(int j =0 ;j<m;j++){
 			StringTokenizer st1 = new StringTokenizer(bf.readLine());
 			int size = Integer.parseInt(st1.nextToken());
 			for(int k =0 ;k<size;k++){
 				int number = Integer.parseInt(st1.nextToken());
 				if (!graph.containsKey(number))
 					graph.put(number,  new ArrayList<Integer>());
 				graph.get(number).add(j+1+n);
 				if (!graph.containsKey(j+1+n))
 					graph.put(j+1+n,  new ArrayList<Integer>());
 				graph.get(j+1+n).add(number);
 				sum+=(costs[j+1+n]+costs[number]);
 				edges.add(new Edge(costs[j+1+n]+costs[number], number, j+1+n));
 			}
 		}
 		Collections.sort(edges, Comparator.comparing(x -> -x.cost));
 		UnionFind uf = new UnionFind(n+m);
 		long ans = 0;
 		for(Edge j : edges){
 			if (uf.get(j.v1) != uf.get(j.v2)){
 				uf.union(j.v1, j.v2);
 				ans+=j.cost;
 			}
 		}
 		out.println(sum-ans);
 		out.close();
 		
 		//cost of an edge is sum of costs of its endpoints
 		
 		
 		
 		
 		
 	}
	static class Edge{
		private long cost;
		private int v1;
		private int v2;
		public Edge(long cost, int v1, int v2) {
			this.cost = cost;
			this.v1 = v1;
			this.v2 = v2;
		}
		
	}
	static class UnionFind{
		private int[] parent;
		private int[] size;
		private int numGroups;
		public UnionFind(int n){
			parent = new int[n+1];
			size = new int[n+1];
			numGroups = n;
			for(int j = 1;j<=n;j++){
				parent[j] = j;
				size[j] = 1;
			}
		}
		public int get(int a){
			if (a!= parent[a])
				parent[a] = get(parent[a]);
			return parent[a];
		}
		public void union(int a, int b){
			int p1 = get(a);
			int p2 = get(b);
			if (size[p1] < size[p2]){
				parent[p1] = p2;
				size[p2]+=size[p1];
			}
			else{
				parent[p2] = p1;
				size[p1]+=size[p2];
			}
			numGroups--;
		}
	}
	
}
 	
 
//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is

 		
 		
 		
 		
	


