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 CF1408E extends PrintWriter {
	CF1408E() { super(System.out, true); }
	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		byte[] bb = new byte[1 << 15]; int i, n;
		byte getc() {
			if (i == n) {
				i = n = 0;
				try { n = in.read(bb); } catch (IOException e) {}
			}
			return i < n ? bb[i++] : 0;
		}
		int nextInt() {
			byte c = 0; while (c <= ' ') c = getc();
			int a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1408E o = new CF1408E(); o.main(); o.flush();
	}

	static final int K = 200000;
	static class E {
		int i, j, w;
		E(int i, int j, int w) {
			this.i = i; this.j = j; this.w = w;
		}
	}
	E[] ee = new E[K];
	int[] ds;
	int find(int i) {
		return ds[i] < 0 ? i : (ds[i] = find(ds[i]));
	}
	boolean join(int i, int j) {
		i = find(i);
		j = find(j);
		if (i == j)
			return false;
		if (ds[i] > ds[j])
			ds[i] = j;
		else {
			if (ds[i] == ds[j])
				ds[i]--;
			ds[j] = i;
		}
		return true;
	}
	void main() {
		int m = sc.nextInt();
		int n = sc.nextInt();
		int[] aa = new int[m];
		int[] bb = new int[n];
		for (int j = 0; j < m; j++)
			aa[j] = sc.nextInt();
		for (int i = 0; i < n; i++)
			bb[i] = sc.nextInt();
		int k = 0;
		for (int j = 0; j < m; j++) {
			int s = sc.nextInt();
			while (s-- > 0) {
				int i = sc.nextInt() - 1;
				ee[k++] = new E(j, m + i, aa[j] + bb[i]);
			}
		}
		Arrays.sort(ee, 0, k, (e, f) -> f.w - e.w);
		ds = new int[m + n]; Arrays.fill(ds, -1);
		long ans = 0;
		for (int h = 0; h < k; h++) {
			E e = ee[h];
			if (!join(e.i, e.j))
				ans += e.w;
		}
		println(ans);
	}
}
