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;
namespace Flandre_Scarlet {
long long I() {
  char c = getchar();
  long long x = 0;
  long long f = 1;
  while (c < '0' or c > '9') f = (c == '-') ? -1 : 1, c = getchar();
  while (c >= '0' and c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return ((f == 1) ? x : -x);
}
template <typename T>
void Rd(T& arg) {
  arg = I();
}
template <typename T, typename... Types>
void Rd(T& arg, Types&... args) {
  arg = I();
  Rd(args...);
}
void RA(long long* p, long long n) {
  for (long long i = 1; i <= n; ++i) *p = I(), ++p;
}
struct edge {
  long long u, v, w;
};
vector<edge> E;
void Add2(long long u, long long v, long long w) {
  E.push_back((edge){u, v, w});
}
long long n, m, a[1000006], b[1000006];
vector<long long> v[1000006];
void Input() {
  Rd(m, n);
  RA(a + 1, m);
  RA(b + 1, n);
  for (long long i = 1; i <= m; ++i) {
    long long k = I();
    for (long long j = 1; j <= k; ++j) v[i].push_back(I());
  }
}
long long color[1000006], vertex[1000006], tot = 0;
class DSU {
 public:
  long long Fa[1000006], Cnt[1000006];
  void clear() {
    for (long long i = 1; i <= 1000006 - 1; ++i) Fa[i] = i, Cnt[i] = 1;
  }
  long long Find(long long u) { return u == Fa[u] ? u : Fa[u] = Find(Fa[u]); }
  bool Merge(long long u, long long v) {
    u = Find(u), v = Find(v);
    if (u == v) return false;
    if (Cnt[u] > Cnt[v]) swap(u, v);
    Fa[u] = v;
    Cnt[v] += Cnt[u];
    return true;
  }
  DSU() { clear(); }
} D;
bool cmp_w(edge a, edge b) { return a.w > b.w; }
long long Kruskal() {
  sort(E.begin(), E.end(), cmp_w);
  long long ans = 0;
  for (auto e : E)
    if (D.Merge(e.u, e.v)) ans += e.w;
  return ans;
}
void Soviet() {
  for (long long i = 1; i <= n; ++i) vertex[i] = ++tot;
  for (long long i = 1; i <= m; ++i) color[i] = ++tot;
  long long sum = 0;
  for (long long i = 1; i <= m; ++i)
    for (auto x : v[i]) sum += a[i] + b[x];
  for (long long i = 1; i <= m; ++i)
    for (auto x : v[i]) Add2(vertex[x], color[i], a[i] + b[x]);
  long long ans = Kruskal();
  printf("%lld\n", sum - ans);
}
void IsMyWife() {
  Input();
  Soviet();
}
}  // namespace Flandre_Scarlet
int main() {
  int fuck_the_network;
  Flandre_Scarlet::IsMyWife();
  getchar();
  return 0;
}
