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.
from sys import stdin, stdout

edges = []
parent = []
height = []
    
def setOf(v):
    p = parent[v]
    if(p == v):
        return p
    return setOf(p)

def merge(v,w):
    if(height[v] > height[w]):
        parent[w] = v
    elif(height[w] > height[v]):
        parent[v] = w
    else:
        parent[w] = v
        height[v] +=1


def Kruskal(n):
    global edges
    edges.sort(key= lambda x: x[0])
    
    global parent
    parent = [i for i in range(n)]
    global height
    height = [0]*n
    mst_cost = 0

    for edge in edges:
        root_a = setOf(edge[1])
        root_b = setOf(edge[2])
        if(root_a != root_b):
            mst_cost -= edge[0]
            merge(root_a, root_b)

    return mst_cost


def avoid_rainbow_cycles(total,n):
    forrest_cost = Kruskal(n)
    stdout.write("{}\n".format(total - forrest_cost))

#input
#line1: m(number of sets), n (number of vertices)
#line2: m integers a1, a2, ..., am (Seti:ai coins)
#line3: n integers b1, b2, ..., bn (Vertexi:bi coins)
#line4-line 4+m: 
# linei: si(size of the set), A1, A1,..., Asi(vertices of the set) 
def main_method():
    m, n = map(int, stdin.readline().split())
    a = list(map(int, stdin.readline().split()))
    b = list(map(int, stdin.readline().split()))

    global edges
    edges = []#a set with <cost,v1,v2> 
    total = 0
    for i in range(m):
        si = list(map(int, stdin.readline().split()))#elements of set i
        csi = a[i]#cost set i
        setNumber = n + i#desfasaje del set i

        for j in range(1,si[0]+1):
            vertex = si[j] - 1
            edge_cost = csi + b[vertex]
            total += edge_cost
            edges.append((-edge_cost, vertex, setNumber))

#verificar que el total este no exceda el mayor valor que python puede representar
    avoid_rainbow_cycles(total, n + m)

main_method()