#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  long long ans = 0;
  vector<int> a(n);
  vector<int> t(n);
  map<int, vector<int> > mp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  int tm = -100;
  long long tot = 0;
  set<pair<int, int> > s;
  while (true) {
    if (s.empty()) {
      auto it = mp.upper_bound(tm);
      if (it == mp.end()) break;
      tm = it->first;
    } else {
      tm++;
    }
    auto q = mp.find(tm);
    if (q != mp.end()) {
      for (auto i : q->second) {
        pair<int, int> p = {t[i], i};
        s.insert(p);
        tot += t[i];
      }
    }
    auto curr = *s.rbegin();
    s.erase(curr);
    tot -= curr.first;
    ans += tot;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  long long ans = 0;
  vector<int> a(n);
  vector<int> t(n);
  map<int, vector<int> > mp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  int tm = -100;
  long long tot = 0;
  set<pair<int, int> > s;
  while (true) {
    if (s.empty()) {
      auto it = mp.upper_bound(tm);
      if (it == mp.end()) break;
      tm = it->first;
    } else {
      tm++;
    }
    auto q = mp.find(tm);
    if (q != mp.end()) {
      for (auto i : q->second) {
        pair<int, int> p = {t[i], i};
        s.insert(p);
        tot += t[i];
      }
    }
    auto curr = *s.rbegin();
    s.erase(curr);
    tot -= curr.first;
    ans += tot;
  }
  cout << ans << endl;
  return 0;
}

```