Problem: Let's denote the following function f. This function takes an array a of length n and returns an array. Initially the result is an empty array. For each integer i from 1 to n we add element a_i to the end of the resulting array if it is greater than all previous elements (more formally, if a_i > max_{1 ≤ j < i}a_j). Some examples of the function f:

  1. if a = [3, 1, 2, 7, 7, 3, 6, 7, 8] then f(a) = [3, 7, 8]; 
  2. if a = [1] then f(a) = [1]; 
  3. if a = [4, 1, 1, 2, 3] then f(a) = [4]; 
  4. if a = [1, 3, 1, 2, 6, 8, 7, 7, 4, 11, 10] then f(a) = [1, 3, 6, 8, 11]. 



You are given two arrays: array a_1, a_2, ... , a_n and array b_1, b_2, ... , b_m. You can delete some elements of array a (possibly zero). To delete the element a_i, you have to pay p_i coins (the value of p_i can be negative, then you get |p_i| coins, if you delete this element). Calculate the minimum number of coins (possibly negative) you have to spend for fulfilling equality f(a) = b.

Input

The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the length of array a.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the array a.

The third line contains n integers p_1, p_2, ..., p_n (|p_i| ≤ 10^9) — the array p.

The fourth line contains one integer m (1 ≤ m ≤ n) — the length of array b.

The fifth line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ n, b_{i-1} < b_i) — the array b.

Output

If the answer exists, in the first line print YES. In the second line, print the minimum number of coins you have to spend for fulfilling equality f(a) = b.

Otherwise in only line print NO.

Examples

Input


11
4 1 3 3 7 8 7 9 10 7 11
3 5 0 -2 5 3 6 7 8 2 4
3
3 7 10


Output


YES
20


Input


6
2 1 5 3 6 5
3 -9 0 16 22 -14
4
2 3 5 6


Output


NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void split(const string& s, char c, vector<string>& result) {
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) result.push_back(x);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << ' ';
  err(++it, args...);
}
struct FenwickTree {
  vector<long long> bit;
  int n;
  FenwickTree(long long n) {
    this->n = n + 1;
    bit.assign(n + 1, 0LL);
  }
  FenwickTree(vector<long long> a) : FenwickTree(a.size()) {
    for (size_t i = 0; i < a.size(); i++) add(i, a[i]);
  }
  long long sum(int idx) {
    long long ret = 0;
    for (++idx; idx > 0; idx -= idx & -idx) ret += bit[idx];
    return ret;
  }
  long long sum(int l, int r) { return sum(r) - sum(l - 1); }
  void add(int idx, long long val) {
    for (++idx; idx < n; idx += idx & -idx) bit[idx] += val;
  }
  void range_add(int l, int r, long long val) {
    add(l, val);
    add(r + 1, -val);
  }
  long long get(int idx) {
    long long ret = 0;
    for (++idx; idx > 0; idx -= idx & -idx) ret += bit[idx];
    return ret;
  }
  void put(int idx, long long val) {
    long long tmp = this->get(idx);
    this->range_add(idx, idx, -tmp);
    this->range_add(idx, idx, val);
  }
};
void readarray(vector<long long>& a, int size) {
  long long tmp;
  a.clear();
  for (int i = 0; i < size; i++) {
    cin >> tmp;
    a.push_back(tmp);
  }
}
long long min3(long long a, long long b, long long c) {
  return min(min(a, b), c);
}
vector<long long> a, b, p;
map<long long, int> pos;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int m, n;
  cin >> n;
  readarray(a, n);
  readarray(p, n);
  cin >> m;
  readarray(b, m);
  FenwickTree mem(m + 1);
  for (int i = 0; i <= m; i++) mem.put(i, 100000000100000000LL);
  mem.put(0, 0);
  for (int i = 0; i < m; i++) pos[b[i]] = 1 + i;
  for (int i = 0; i < n; i++) {
    long long cur = a[i];
    long long cost = p[i];
    if (pos.find(cur) == pos.end()) {
      if (cost <= 0) {
        mem.add(0, cost);
      } else {
        int last = lower_bound(b.begin(), b.end(), cur) - b.begin();
        mem.range_add(0, last, cost);
      }
    } else {
      long long addcost1 = mem.get(pos[cur] - 1);
      long long addcost2 = mem.get(pos[cur]);
      long long addcost3 = addcost2 + cost;
      long long mincost = min3(addcost1, addcost2, addcost3);
      mem.put(pos[cur], mincost);
      mem.range_add(0, pos[cur] - 1, cost);
      if (cost <= 0) {
        mem.range_add(pos[cur] + 1, m, cost);
      }
    }
  }
  long long ans = mem.get(m);
  if (ans * 10 > 100000000100000000LL)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl << ans << endl;
  }
  return 0;
}