VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.

The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds. 

What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.

Input

The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).

The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).

The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).

Output

Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.

Examples

Input


5
3 7 9 7 8
5 2 5 7 5


Output


6


Input


5
1 2 3 4 5
1 1 1 1 1


Output


0

Note

In the first example, it is possible to find three publications of the second type, which will take 6 seconds.

In the second example, all news categories contain a different number of publications.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int a[n], t[n];
  pair<int, int> pr[n];
  for (int i = 0; i < n; i++) {
    cin >> pr[i].first;
    pr[i].second = i;
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  sort(pr, pr + n);
  multiset<int> st;
  long long int sm = 0;
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    if (i < n - 1 and pr[i + 1].first == pr[i].first) {
      sm += t[pr[i].second];
      st.insert(t[pr[i].second]);
    } else {
      sm += t[pr[i].second];
      st.insert(t[pr[i].second]);
      auto it = st.end();
      it--;
      sm -= *it;
      ans += sm;
      st.erase(it);
      int x = pr[i].first + 1;
      while (i < n - 1 and x < pr[i + 1].first and !st.empty()) {
        it = st.end();
        it--;
        sm -= *it;
        ans += sm;
        st.erase(it);
        x++;
      }
    }
  }
  while (!st.empty()) {
    auto it = st.end();
    it--;
    sm -= *it;
    ans += sm;
    st.erase(it);
  }
  cout << ans << "\n";
  return 0;
}
