#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  priority_queue<pair<int, int> > pq;
  int n;
  cin >> n;
  vector<pair<int, int> > v(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v[i].first = x;
  }
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    v[i].second = t;
  }
  sort(v.begin(), v.end());
  int prev = v[0].first;
  pq.push({v[0].second, v[0].first});
  long long ans = 0;
  for (int i = 1; i < n; i++) {
    int cur = v[i].first;
    while (prev < cur && !pq.empty()) {
      pair<int, int> p = pq.top();
      ans += 1LL * (prev - p.second) * p.first;
      pq.pop();
      ++prev;
    }
    pq.push({v[i].second, v[i].first});
    prev = cur;
  }
  while (!pq.empty()) {
    pair<int, int> p = pq.top();
    ans += 1LL * (prev - p.second) * p.first;
    pq.pop();
    ++prev;
  }
  cout << ans;
  return 0;
}

### 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  priority_queue<pair<int, int> > pq;
  int n;
  cin >> n;
  vector<pair<int, int> > v(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v[i].first = x;
  }
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    v[i].second = t;
  }
  sort(v.begin(), v.end());
  int prev = v[0].first;
  pq.push({v[0].second, v[0].first});
  long long ans = 0;
  for (int i = 1; i < n; i++) {
    int cur = v[i].first;
    while (prev < cur && !pq.empty()) {
      pair<int, int> p = pq.top();
      ans += 1LL * (prev - p.second) * p.first;
      pq.pop();
      ++prev;
    }
    pq.push({v[i].second, v[i].first});
    prev = cur;
  }
  while (!pq.empty()) {
    pair<int, int> p = pq.top();
    ans += 1LL * (prev - p.second) * p.first;
    pq.pop();
    ++prev;
  }
  cout << ans;
  return 0;
}

```