#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > a;
multiset<long long> s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x, t, m, sum = 0, curr, ans = 0;
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> a[i].second;
  }
  sort(a.begin(), a.end());
  s.insert(a[0].second);
  curr = a[0].first;
  sum = a[0].second;
  bool flag = false;
  for (int i = 1; i < n; i++) {
    if (a[i].first != curr) {
      if (s.size() == 0) {
        curr = a[i].first;
      } else {
        sum -= *prev(s.end());
        s.erase(prev(s.end()));
        ans += sum;
        curr++;
      }
      i--;
      continue;
    }
    while (a[i].first == curr) {
      sum += a[i].second;
      s.insert(a[i].second);
      i++;
      if (i == n) {
        flag = true;
        break;
      }
    }
    if (flag) {
      break;
    }
    i--;
  }
  while (s.size() != 0) {
    sum -= *prev(s.end());
    s.erase(prev(s.end()));
    ans += sum;
    curr++;
  }
  cout << ans;
  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;
vector<pair<long long, long long> > a;
multiset<long long> s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x, t, m, sum = 0, curr, ans = 0;
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> a[i].second;
  }
  sort(a.begin(), a.end());
  s.insert(a[0].second);
  curr = a[0].first;
  sum = a[0].second;
  bool flag = false;
  for (int i = 1; i < n; i++) {
    if (a[i].first != curr) {
      if (s.size() == 0) {
        curr = a[i].first;
      } else {
        sum -= *prev(s.end());
        s.erase(prev(s.end()));
        ans += sum;
        curr++;
      }
      i--;
      continue;
    }
    while (a[i].first == curr) {
      sum += a[i].second;
      s.insert(a[i].second);
      i++;
      if (i == n) {
        flag = true;
        break;
      }
    }
    if (flag) {
      break;
    }
    i--;
  }
  while (s.size() != 0) {
    sum -= *prev(s.end());
    s.erase(prev(s.end()));
    ans += sum;
    curr++;
  }
  cout << ans;
  return 0;
}

```