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 comment(linker, "/stack:512000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937_64 gen(time(nullptr));
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
ll mod = 1e9 + 7;
namespace {
ll mul(ll a, ll b) {
  ll val = a * b - (ll)((ld)a * b / mod) * mod;
  if (val < 0) val += mod;
  if (val >= mod) val -= mod;
  return val;
}
ll poww(ll a, ll b) {
  ll val = 1;
  a %= mod;
  while (b > 0) {
    if (b % 2) val = mul(a, val);
    a = mul(a, a);
    b >>= 1;
  }
  return val % mod;
}
ll inv(ll a) { return poww(a, mod - 2); }
ll gcd(ll a, ll b) { return (a == 0 ? b : gcd(b % a, a)); }
ll gcdex(ll a, ll b, ll& x, ll& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  ll x1, y1;
  ll d = gcdex(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
ll inv_eu(ll a, ll m) {
  ll x, y;
  ll g = gcdex(a, m, x, y);
  if (g != 1)
    assert(0);
  else {
    x = (x % m + m) % m;
    return x;
  }
}
}  // namespace
ll const maxn = 3e5 + 5;
int par[maxn], sz[maxn];
void dsu_make(int u) {
  par[u] = u;
  sz[u] = 1;
}
int dsu_get(int u) { return (par[u] == u ? u : par[u] = dsu_get(par[u])); }
void dsu_merge(int u, int v) {
  u = dsu_get(u), v = dsu_get(v);
  if (u == v) return;
  if (sz[u] < sz[v]) swap(u, v);
  par[v] = u;
  sz[u] += sz[v];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m, n;
  cin >> m >> n;
  vector<int> a(m), b(n);
  for (int i = 0; i < m; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  vector<pair<int, pair<int, int>>> kek;
  ll ans1 = 0;
  for (int i = 0; i < m; i++) {
    int s;
    cin >> s;
    for (int j = 0; j < s; j++) {
      int x;
      cin >> x;
      x--;
      kek.push_back({a[i] + b[x], {i, x}});
      ans1 += (a[i] + b[x]);
    }
  }
  ll ans = 0;
  sort(kek.begin(), kek.end());
  reverse(kek.begin(), kek.end());
  vector<vector<int>> g(n);
  for (int i = 0; i < m; i++) {
    dsu_make(i);
  }
  for (auto p : kek) {
    int x = p.first;
    int i = p.second.first, j = p.second.second;
    bool f = true;
    for (int ii : g[j]) {
      if (dsu_get(ii) == dsu_get(i)) {
        f = false;
        break;
      } else {
        break;
      }
    }
    if (f) {
      ans += x;
      for (int ii : g[j]) {
        dsu_merge(ii, i);
        break;
      }
      g[j].emplace_back(i);
    }
  }
  cout << ans1 - ans << endl;
  return 0;
}
