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;
const long long int MOD = 1000000007;
const long long int INF = 2147483647;
const long long int LINF = 9223372036854775807;
struct att {
  long long int a;
  long long int t;
  bool operator<(const att &another) const { return a < another.a; };
};
int main() {
  int n;
  cin >> n;
  vector<att> at(n);
  for (int j = 0; j < n; j++) cin >> at[j].a;
  for (int j = 0; j < n; j++) cin >> at[j].t;
  sort(at.begin(), at.end());
  priority_queue<pair<long long int, long long int> > q;
  long long int ans = 0;
  for (int j = 0; j < n - 1; j++) {
    if (at[j].a == at[j + 1].a) {
      q.push({at[j].t, at[j].a});
    } else if (!q.empty()) {
      if (q.top().first > at[j].t) {
        ans += q.top().first * (at[j].a - q.top().second);
        q.pop();
        q.push({at[j].t, at[j].a});
      }
      for (int k = 1; 1; k++) {
        if (at[j].a + k < at[j + 1].a) {
          ans += q.top().first * (at[j].a + k - q.top().second);
          q.pop();
          if (q.empty()) break;
        } else {
          break;
        }
      }
    }
  }
  if (!q.empty()) {
    if (q.top().first > at[n - 1].t) {
      ans += q.top().first * (at[n - 1].a - q.top().second);
      q.pop();
      q.push({at[n - 1].t, at[n - 1].a});
    }
  }
  for (int k = 1; 1; k++) {
    if (!q.empty()) {
      ans += q.top().first * (at[n - 1].a + k - q.top().second);
      q.pop();
    } else {
      break;
    }
  }
  cout << ans << endl;
  return 0;
}
