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;
void solve();
bool svp(const pair<long long, long long> &a,
         const pair<long long, long long> &b) {
  if (a.first != b.first) return (a.first < b.first);
  return (a.second > b.second);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
}
void solve() {
  long long n = 0, ans = 0, s = 0;
  cin >> n;
  pair<long long, long long> b[n];
  for (int i = 0; i < n; i++) cin >> b[i].first;
  for (int i = 0; i < n; i++) cin >> b[i].second;
  sort(b, b + n, svp);
  multiset<long long, greater<long long>> ms;
  ms.insert(b[0].second);
  s = b[0].second;
  for (int i = 1; i <= n - 1; i++) {
    if (b[i].first == b[i - 1].first) {
      s += b[i].second;
      ms.insert(b[i].second);
    } else {
      long long i1 = 0;
      for (auto it = ms.begin();
           it != ms.end() && (i1 < b[i].first - b[i - 1].first);) {
        ans += i1 * (*it);
        auto it1 = it;
        it++;
        if (b[i - 1].first + i1 < b[i].first) {
          s -= (*it1);
          ms.erase(it1);
          i1++;
        }
      }
      ans += s * i1;
      s += b[i].second;
      ms.insert(b[i].second);
    }
  }
  int i1 = 0;
  for (auto j : ms) {
    ans += i1 * j;
    i1++;
  }
  cout << ans << endl;
}
