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.io.*; import java.util.*;
public class E{
    public static class Edge implements Comparable<Edge>{
        int v1; int v2; int val;
        public Edge(int a, int b, int c){
            this.v1=a; this.v2=b; this.val=c;
        }
        public int compareTo(Edge other){
            if(this.val>other.val)return -1;
            return 1;
        }
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        int M=Integer.parseInt(st.nextToken());
        int N=Integer.parseInt(st.nextToken());
        st=new StringTokenizer(br.readLine());
        int[] a=new int[M];
        for (int i = 0; i < M; i++) {
            a[i]=Integer.parseInt(st.nextToken());
        }
        st=new StringTokenizer(br.readLine());
        int[] b=new int[N];
        for (int i = 0; i < N; i++) {
            b[i]=Integer.parseInt(st.nextToken());
        }
        int s=0; int t=0; long ans=0;
        ArrayList<Edge> arl=new ArrayList<>();
        for (int i = 0; i < M; i++) {
            st=new StringTokenizer(br.readLine());
            s=Integer.parseInt(st.nextToken());
            for (int j = 0; j < s; j++) {
                t=Integer.parseInt(st.nextToken())-1;
                arl.add(new Edge(i,M+t,a[i]+b[t]));
                ans+=(a[i]+b[t]);
            }
        }
        Collections.sort(arl);
        int[] rank=new int[N+M];
        int[] par=new int[N+M];
        for (int i = 1; i < N+M; i++) {
            par[i]=i;
        }
        int done=0; int pt=0;
        int x,y=0; 
        while(pt<arl.size()){//Maximum spanning tree
            Edge e=arl.get(pt);
            x=find(e.v1,par); y=find(e.v2,par);
            if(x!=y){
                done++; ans-=e.val;
                if(rank[x]<rank[y]){
                    par[x]=y;
                }else if(rank[x]==rank[y]){
                    rank[y]++; par[x]=y;
                }else{
                    par[y]=x;
                }
                if(done==M+N-1){
                    break;
                }
            }
            
            pt++;
        }
        System.out.println(ans);
    }
    public static int find(int x, int[] p){
        if(p[x]==x)return x;
        int ans=find(p[x],p); p[x]=ans; return ans;
    }
}
