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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long MOD) {
  if (a == 0ll) {
    return 0ll;
  }
  a %= MOD;
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return ans;
}
long long poww(long long a, long long b) {
  if (a == 0ll) {
    return 0ll;
  }
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a;
    }
    a = a * a;
    b >>= 1;
  }
  return ans;
}
void Pv(const vector<int> &V) {
  int Len = int(V.size());
  for (int i = 0; i < Len; ++i) {
    printf("%d", V[i]);
    if (i != Len - 1) {
      printf(" ");
    } else {
      printf("\n");
    }
  }
}
void Pvl(const vector<long long> &V) {
  int Len = int(V.size());
  for (int i = 0; i < Len; ++i) {
    printf("%lld", V[i]);
    if (i != Len - 1) {
      printf(" ");
    } else {
      printf("\n");
    }
  }
}
inline long long readll() {
  long long tmp = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar();
  return tmp * fh;
}
inline int readint() {
  int tmp = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar();
  return tmp * fh;
}
void pvarr_int(int *arr, int n, int strat = 1) {
  if (strat == 0) {
    n--;
  }
  for (int i = strat; i <= n; i++) {
    printf("%d%c", arr[i], i == n ? '\n' : ' ');
  }
}
void pvarr_LL(long long *arr, int n, int strat = 1) {
  if (strat == 0) {
    n--;
  }
  for (int i = strat; i <= n; i++) {
    printf("%lld%c", arr[i], i == n ? '\n' : ' ');
  }
}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
int n, m;
long long a[maxn], b[maxn];
struct node {
  int f, t;
  long long val;
  node() {}
  node(int ff, int tt, long long vv) {
    f = ff;
    t = tt;
    val = vv;
  }
  bool operator<(const node &bb) const { return val > bb.val; }
};
std::vector<node> v;
int far[maxn];
void dsu_init(int n) {
  for (int i = 0; i <= n; i++) {
    far[i] = i;
  }
}
int findpar(int x) {
  if (x == far[x]) {
    return x;
  } else {
    return far[x] = findpar(far[x]);
  }
}
void mg(int x, int y) {
  x = findpar(x);
  y = findpar(y);
  if (x == y) return;
  far[x] = y;
}
long long maxst() {
  dsu_init(n + m);
  sort((v).begin(), (v).end());
  long long res = 0ll;
  for (auto e : v) {
    int x = e.f;
    int y = e.t;
    x = findpar(x);
    y = findpar(y);
    if (x != y) {
      res += e.val;
      far[x] = y;
    }
  }
  return res;
}
int main() {
  m = readint();
  n = readint();
  for (int i = 1; i <= m; i++) {
    a[i] = readll();
  }
  for (int i = 1; i <= n; i++) {
    b[i] = readll();
  }
  long long ans = 0ll;
  for (int i = 1; i <= m; i++) {
    int k = readint();
    while (k--) {
      int x = readint();
      ans += a[i] + b[x];
      v.push_back(node(i, m + x, a[i] + b[x]));
      v.push_back(node(m + x, i, a[i] + b[x]));
    }
  }
  printf("%lld\n", ans - maxst());
  return 0;
}
