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.
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long maxn = 1e6 + 100;
long long n, m, k, t;
vector<long long> X[maxn];
vector<long long> Y[maxn];
long long A[maxn];
long long B[maxn];
bool cmpa(long long i, long long j) { return A[i] > A[j]; }
bool cmpb(long long i, long long j) { return B[i] > B[j]; }
long long P[maxn];
inline long long getp(long long i) {
  if (P[i] == i) {
    return i;
  }
  long long ans = getp(P[i]);
  P[i] = ans;
  return ans;
}
inline void mrg(long long i, long long j) {
  i = getp(i);
  j = getp(j);
  if (i != j) {
    P[j] = i;
  }
}
bool cmpp(pair<long long, long long> a, pair<long long, long long> b) {
  return A[a.first] + B[a.second] > A[b.first] + B[b.second];
}
long long R[maxn];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long q, w, e, a, b, c;
  cin >> m >> n;
  vector<long long> I1, I2;
  for (q = 0; q < maxn; q++) {
    P[q] = q;
  }
  for (q = 0; q < m; q++) {
    I1.push_back(q);
    cin >> A[q];
  }
  for (q = 0; q < n; q++) {
    I2.push_back(q);
    cin >> B[q];
  }
  vector<pair<long long, long long> > PP;
  for (q = 0; q < m; q++) {
    cin >> k;
    for (w = 0; w < k; w++) {
      cin >> a;
      a--;
      PP.push_back(make_pair(q, a));
      X[a].push_back(q);
      Y[q].push_back(a);
    }
  }
  sort(I1.begin(), I1.end(), cmpa);
  sort(I2.begin(), I2.end(), cmpb);
  for (q = 0; q < m; q++) {
    sort(Y[q].begin(), Y[q].end(), cmpb);
  }
  for (q = 0; q < n; q++) {
    sort(X[q].begin(), X[q].end(), cmpa);
    if (!X[q].empty()) {
      R[q] = X[q][0];
    }
  }
  long long ans = 0;
  sort(PP.begin(), PP.end(), cmpp);
  for (auto p : PP) {
    long long v1 = p.first, v2 = R[p.second];
    if (getp(v1) != getp(v2) || v1 == v2) {
      mrg(v1, v2);
    } else {
      ans += A[p.first];
      ans += B[p.second];
    }
  }
  cout << ans;
  return 0;
}
