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 powArray(long long[], long long = 2, int = 62);
vector<int> getBin(long long, int = 62);
long long getDec(vector<int>);
int numOfOnes(long long n);
int logN(long long, long long = 2);
long long inf = 1e16, a[500500], b[500500], p[500500], n, m, seg[5005000];
pair<long long, long long> dp[500500];
bool can[500500];
map<pair<long long, pair<long long, long long>>, long long> querys, querysAns;
vector<int> numbers[500500];
void build() {
  for (int i = n - 1; i > 0; --i) seg[i] = seg[i << 1] + seg[i << 1 | 1];
}
void update(int p, long long first) {
  for (seg[p += n] = first; p > 1; p >>= 1) seg[p >> 1] = seg[p] + seg[p ^ 1];
}
long long query(int l, int r) {
  r++;
  long long res = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      res += seg[l];
      l++;
    }
    if (r & 1) {
      r--;
      res += seg[r];
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  for (int i = 0; i < 500500; i++) {
    dp[i].first = inf;
    dp[i].second = -1;
  }
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> p[i];
  for (int i = 0; i < n; i++) {
    numbers[a[i]].push_back(i);
    seg[i + n] = p[i];
  }
  build();
  cin >> m;
  for (int i = 1; i <= m; i++) cin >> b[i];
  m++;
  can[0] = true;
  for (int i = 0; i < n; i++) {
    int j = lower_bound(b, b + m, a[i]) - b;
    if (b[j] == a[i]) {
      if (can[b[j - 1]]) {
        querys[{b[j - 1], {dp[b[j - 1]].second + 1, i}}] = inf;
        if (can[b[j]]) {
          querys[{b[j], {dp[b[j]].second + 1, i}}] = inf;
        }
        can[b[j]] = true;
        dp[b[j]].second = i;
      }
    }
  }
  querys[{b[m - 1], {dp[b[m - 1]].second + 1, n - 1}}] = inf;
  long long last = 0;
  for (auto first : querys) {
    for (int i = last; i < first.first.first; i++) {
      for (auto second : numbers[i])
        if (p[second] > 0) update(second, 0);
    }
    last = first.first.first;
    querysAns[first.first] =
        query(first.first.second.first, first.first.second.second);
  }
  for (int i = 0; i < 500500; i++) {
    dp[i].first = inf;
    dp[i].second = -1;
    can[i] = false;
  }
  can[0] = true;
  dp[0].first = 0;
  for (int i = 0; i < n; i++) {
    int j = lower_bound(b, b + m, a[i]) - b;
    if (b[j] == a[i]) {
      if (can[b[j - 1]]) {
        if (can[b[j]]) {
          dp[b[j]].first =
              min(dp[b[j]].first + querysAns[{b[j], {dp[b[j]].second + 1, i}}] -
                      p[i],
                  dp[b[j]].first + querysAns[{b[j], {dp[b[j]].second + 1, i}}]);
        }
        dp[b[j]].first =
            min(dp[b[j - 1]].first +
                    querysAns[{b[j - 1], {dp[b[j - 1]].second + 1, i}}] - p[i],
                dp[b[j]].first);
        can[b[j]] = true;
        dp[b[j]].second = i;
      }
    }
  }
  if (can[b[m - 1]])
    cout << "YES\n"
         << dp[b[m - 1]].first +
                querysAns[{b[m - 1], {dp[b[m - 1]].second + 1, n - 1}}];
  else
    cout << "NO\n";
}
void powArray(long long array[], long long base, int size) {
  long long current = 1;
  for (int i = 0; i <= size; i++) {
    array[i] = current;
    current *= base;
  }
}
vector<int> getBin(long long n, int size) {
  vector<int> result;
  for (int i = 0; i < size; i++) {
    result.push_back(n % 2);
    n /= 2;
  }
  reverse(result.begin(), result.end());
  return result;
}
long long getDec(vector<int> n) {
  long long result = 0;
  long long p[100];
  powArray(p);
  for (int i = 0; i < n.size(); i++) {
    result += p[n.size() - i - 1] * n[i];
  }
  return result;
}
int numOfOnes(long long n) {
  vector<int> nBin = getBin(n);
  int result = 0;
  for (auto first : nBin) result += first;
  return result;
}
int logN(long long n, long long base) {
  long long result = 0;
  while (n) {
    result++;
    n /= base;
  }
  return result - 1;
}