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;
template <class T>
struct node {
  T val;
  T lazy;
  int l, r;
  node* left;
  node* right;
  node(int l, int r) {
    this->l = l;
    this->r = r;
    this->left = nullptr;
    this->right = nullptr;
  }
};
template <class T, int SZ>
struct segtree {
  T combIdentity = 1e17;
  T comb(T l, T r) { return min(l, r); }
  T pushIdentity = 0;
  void push(node<T>* n) {
    n->val += n->lazy;
    if (n->l != n->r) {
      n->left->lazy += n->lazy;
      n->right->lazy += n->lazy;
    }
    n->lazy = pushIdentity;
  }
  node<T>* root;
  segtree() {
    int ub = 1;
    while (ub < SZ) ub *= 2;
    root = new node<T>(0, ub - 1);
    root->val = pushIdentity;
    root->lazy = pushIdentity;
  }
  void propagate(node<T>* n) {
    if (n->l != n->r) {
      int mid = ((n->l) + (n->r)) / 2;
      if (n->left == nullptr) {
        n->left = new node<T>(n->l, mid);
        n->left->val = pushIdentity;
        n->left->lazy = pushIdentity;
      }
      if (n->right == nullptr) {
        n->right = new node<T>(mid + 1, n->r);
        n->right->val = pushIdentity;
        n->right->lazy = pushIdentity;
      }
    }
    push(n);
  }
  void updN(node<T>* n, int i1, int i2, T val) {
    propagate(n);
    if (i2 < n->l || i1 > n->r) return;
    if (i1 <= n->l && i2 >= n->r) {
      n->lazy = val;
      return;
    }
    updN(n->left, i1, i2, val);
    updN(n->right, i1, i2, val);
    n->val = comb(n->left->val, n->right->val);
  }
  void upd(int i1, int i2, T val) { updN(root, i1, i2, val); }
  T queryN(node<T>* n, int i1, int i2) {
    propagate(n);
    if (i2 < n->l || i1 > n->r) return combIdentity;
    if (n->l >= i1 && n->r <= i2) return n->val;
    T a = combIdentity;
    if (n->left != nullptr) a = comb(a, queryN(n->left, i1, i2));
    if (n->right != nullptr) a = comb(a, queryN(n->right, i1, i2));
    return a;
  }
  T query(int i1, int i2) { return queryN(root, i1, i2); }
};
const int MAX_N = 500500;
int n;
int arr[MAX_N];
long long cost[MAX_N];
int k;
int des[MAX_N];
map<int, int> desIdx;
segtree<long long, MAX_N> dp;
void setVal(int idx, long long val) {
  dp.upd(idx, idx, -dp.query(idx, idx) + val);
  assert(dp.query(idx, idx) == val);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) cin >> cost[i];
  cin >> k;
  for (int i = 0; i < k; i++) cin >> des[i];
  for (int i = 0; i < k; i++) desIdx[des[i]] = i + 1;
  dp.upd(0, k, 1e16);
  setVal(0, cost[0]);
  if (arr[0] == des[0]) setVal(desIdx[arr[0]], 0);
  for (int i = 1; i < n; i++) {
    if (arr[i] < des[0]) {
      dp.upd(1, k, min(cost[i], 0LL));
    } else if (desIdx.count(arr[i])) {
      int idx = desIdx[arr[i]];
      long long val = min(dp.query(idx - 1, idx - 1),
                          dp.query(idx, idx) + min(cost[i], 0LL));
      if (val <= 1e15) setVal(idx, val);
      if (1 <= idx - 1) dp.upd(1, idx - 1, cost[i]);
      if (idx + 1 <= k && cost[i] < 0) dp.upd(idx + 1, k, cost[i]);
    } else {
      auto it = desIdx.upper_bound(arr[i]);
      if (it == desIdx.end()) {
        dp.upd(1, k, cost[i]);
      } else {
        int idx = (*it).second;
        if (1 <= idx - 1) dp.upd(1, idx - 1, cost[i]);
        if (cost[i] < 0) dp.upd(idx, k, cost[i]);
      }
    }
    dp.upd(0, 0, cost[i]);
  }
  long long res = dp.query(k, k);
  if (abs(res) > (long long)MAX_N * 1e9) {
    return cout << "NO" << '\n', 0;
    ;
  }
  cout << "YES"
       << "\n";
  return cout << res << '\n', 0;
  ;
}