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.*;
 
 
//Sub_To_Errichto
public class Main {
    static final long MOD = 1000000007L;
    //static final long MOD = 998244353L;
    static final int INF = 50000000;
    static final int NINF = -500000000;
    
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter pw = new PrintWriter(System.out);
        
    	int M = sc.ni();
    	int N = sc.ni();
    	long[] A = new long[M];
    	for (int i = 0; i < M; i++)
    		A[i] = sc.nl();
    	long[] B = new long[N];
    	for (int i = 0; i < N; i++)
    		B[i] = sc.nl();
    	
    	ArrayList<Edge> edges = new ArrayList<Edge>();
    	long tot = 0;
    	for (int a = 0; a < M; a++) {
    		int S = sc.ni();
    		for (int j = 0; j < S; j++) {
    			int b = sc.ni()-1;
    			long w = A[a]+B[b];
    			tot += w;
    			edges.add(new Edge(a,M+b,w));
    		}
    	}
    	Collections.sort(edges);
    	DisjointSetUnion dsu = new DisjointSetUnion(N+M);
    	long ans = 0;
    	for (Edge e: edges) {
    		if (!dsu.connected(e.a,e.b)) {
    			dsu.connect(e.a, e.b);
    			ans += e.w;
    		}
    	}
    	pw.println(tot-ans);
        pw.close();
    }
    
    //implementation of union find data structure
    static class DisjointSetUnion {
    	public int[] parent;
    	public int[] weight;
    	public int count;

    	public DisjointSetUnion(int N) {
    		count = N;
    		parent = new int[N];
    		for (int i = 0; i < N; i++)
    			parent[i] = i;
    		weight = new int[N];
    		Arrays.fill(weight,1);
    	}

    	//"find"
    	public int root(int p) {
    		while (p != parent[p]) {
    			p = parent[p];
    		}
    		return p;
    	}

    	//"union"
    	public void connect(int p, int q) {
    		p = root(p);
    		q = root(q);
    		if (p==q)
    			return;
    		if (weight[p] < weight[q]) {
    			parent[p] = q;
    			weight[q] += weight[p];
    		} else {
    			parent[q] = p;
    			weight[p] += weight[q];
    		}
    		count--;
    	}

    	public boolean connected(int p, int q) {
    		return root(p) == root(q);
    	}
    }
    
    static class Edge implements Comparable<Edge> {
    	int a;
    	int b;
    	long w;
    	public Edge(int a, int b, long w) {
    		this.a=a;
    		this.b=b;
    		this.w=w;
    	}
    	
    	public int compareTo(Edge e) {
    		if (w < e.w)
    			return 1;
    		else if (w > e.w)
    			return -1;
    		else
    			return 0;
    	}
    	
    	public String toString() {
    		return "[" + a + "," + b + "," + w + "]";
    	}
    }
    
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int ni() {
            return Integer.parseInt(next());
        }
 
        long nl() {
            return Long.parseLong(next());
        }
 
        double nd() {
            return Double.parseDouble(next());
        }
 
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}