#include <bits/stdc++.h>
using namespace std;
struct solver {
  long long key, sum = 0;
  priority_queue<long long> pq;
  void add(long long x) {
    pq.push(x);
    sum += x;
  }
  long long extract_max() {
    long long x = pq.top();
    pq.pop();
    sum -= x;
    return x;
  }
  bool empty() { return pq.empty(); }
};
int32_t main() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i].first;
  for (long long i = 0; i < n; ++i) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long ans = 0;
  solver s;
  long long a_idx = 0;
  s.key = a[a_idx].first;
  while (a_idx < n) {
    if (s.empty()) {
      s.key = a[a_idx].first;
    }
    while (a_idx < n && a[a_idx].first == s.key) {
      s.add(a[a_idx].second);
      ++a_idx;
    }
    s.extract_max();
    ans += s.sum;
    if (!s.empty()) ++s.key;
  }
  while (!s.empty()) {
    s.extract_max();
    ans += s.sum;
  }
  cout << ans << endl;
}

### 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;
struct solver {
  long long key, sum = 0;
  priority_queue<long long> pq;
  void add(long long x) {
    pq.push(x);
    sum += x;
  }
  long long extract_max() {
    long long x = pq.top();
    pq.pop();
    sum -= x;
    return x;
  }
  bool empty() { return pq.empty(); }
};
int32_t main() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i].first;
  for (long long i = 0; i < n; ++i) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long ans = 0;
  solver s;
  long long a_idx = 0;
  s.key = a[a_idx].first;
  while (a_idx < n) {
    if (s.empty()) {
      s.key = a[a_idx].first;
    }
    while (a_idx < n && a[a_idx].first == s.key) {
      s.add(a[a_idx].second);
      ++a_idx;
    }
    s.extract_max();
    ans += s.sum;
    if (!s.empty()) ++s.key;
  }
  while (!s.empty()) {
    s.extract_max();
    ans += s.sum;
  }
  cout << ans << endl;
}

```