#include <bits/stdc++.h>
using namespace std;
bool comp(pair<int, int> x, pair<int, int> y) {
  if (x.first != y.first) return x.first < y.first;
  return x.second > y.second;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int> > x(n);
  for (int i = 0; i < n; i++) cin >> x[i].first;
  for (int i = 0; i < n; i++) cin >> x[i].second;
  sort(x.begin(), x.end(), comp);
  priority_queue<int> q;
  long long res = 0;
  long long last = 0, nowsum = 0;
  for (int i = 0; i < n; i++) {
    while (!q.empty() && x[i].first > last) {
      nowsum -= q.top();
      q.pop();
      res += nowsum;
      last++;
    }
    last = x[i].first;
    q.push(x[i].second);
    nowsum += x[i].second;
  }
  while (!q.empty()) {
    nowsum -= q.top();
    q.pop();
    res += nowsum;
    last++;
  }
  cout << res;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool comp(pair<int, int> x, pair<int, int> y) {
  if (x.first != y.first) return x.first < y.first;
  return x.second > y.second;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int> > x(n);
  for (int i = 0; i < n; i++) cin >> x[i].first;
  for (int i = 0; i < n; i++) cin >> x[i].second;
  sort(x.begin(), x.end(), comp);
  priority_queue<int> q;
  long long res = 0;
  long long last = 0, nowsum = 0;
  for (int i = 0; i < n; i++) {
    while (!q.empty() && x[i].first > last) {
      nowsum -= q.top();
      q.pop();
      res += nowsum;
      last++;
    }
    last = x[i].first;
    q.push(x[i].second);
    nowsum += x[i].second;
  }
  while (!q.empty()) {
    nowsum -= q.top();
    q.pop();
    res += nowsum;
    last++;
  }
  cout << res;
}

```