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>
using namespace std;
constexpr long long MOD = 1000000007;
constexpr int INF = 0x7f7f7f7f;
constexpr double EPS = 1e-8;
long long fastPow(long long num, long long exp, long long modNum) {
  if (exp == 0) return 1;
  if (exp == 1)
    return num;
  else {
    long long half = fastPow(num, exp / 2, modNum) % modNum;
    if (exp % 2 == 1) {
      return (((half * half) % modNum) * num) % modNum;
    } else {
      return (half * half) % modNum;
    }
  }
}
long long fastPow(long long num, long long exp) {
  if (exp == 0) return 1;
  if (exp == 1)
    return num;
  else {
    long long half = fastPow(num, exp / 2);
    if (exp % 2 == 1) {
      return (((half * half)) * num);
    } else {
      return (half * half);
    }
  }
}
long long getReverse(long long y, long long modNum) {
  return fastPow(y, modNum - 2, modNum) % modNum;
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  return (a % b == 0) ? b : gcd(b, a % b);
}
long long find(long long k, vector<long long> &F) {
  return F[k] == k ? k : F[k] = find(F[k], F);
}
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  const long long MOD = 1e9 + 7;
  int T = 1;
  for (int(I) = (0); (I) <= ((T)-1); (I)++) {
    long long m, n;
    cin >> m >> n;
    vector<long long> a(m), b(n + 1);
    for (int(i) = (0); (i) <= ((m)-1); (i)++) {
      cin >> a[i];
    }
    for (int(i) = (1); (i) <= (n); (i)++) {
      cin >> b[i];
    }
    vector<tuple<long long, long long, long long>> data;
    for (int(i) = (0); (i) <= ((m)-1); (i)++) {
      long long sz, val;
      cin >> sz;
      for (int(j) = (0); (j) <= ((sz)-1); (j)++) {
        cin >> val;
        long long cost = a[i] + b[val];
        data.emplace_back(cost, i, val);
      }
    }
    sort((data).begin(), (data).end(),
         greater<tuple<long long, long long, long long>>());
    long long ans = 0;
    vector<long long> tags(m, -1);
    vector<long long> F(n + 1);
    for (int(i) = (0); (i) <= ((n + 1) - 1); (i)++) {
      F[i] = i;
    }
    for (auto tu : data) {
      long long cost, group, val;
      tie(cost, group, val) = tu;
      if (tags[group] == -1) {
        tags[group] = val;
      } else {
        if (find(val, F) == find(tags[group], F)) {
          ans += cost;
        } else {
          F[find(val, F)] = find(tags[group], F);
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
