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;
struct Obj {
  int val, time;
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  vector<Obj> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].val;
  }
  for (int i = 0; i < n; ++i) {
    cin >> a[i].time;
  }
  sort(a.begin(), a.end(), [](const Obj &left, const Obj &right) {
    if (left.val == right.val) {
      return left.time < right.time;
    }
    return left.val < right.val;
  });
  priority_queue<pair<int, int> > pq;
  int next;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    next = (i == n - 1 ? INT_MAX : a[i + 1].val);
    pq.push({a[i].time, a[i].val});
    int j = a[i].val;
    while (!pq.empty() && j < next) {
      auto cur = pq.top();
      pq.pop();
      ans += 1LL * cur.first * (j - cur.second);
      j++;
    }
  }
  cout << ans;
  return 0;
}
