#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> arr;
  arr.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i].first;
    arr[i].second = i;
  }
  vector<int> t;
  t.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  sort(arr.begin(), arr.end());
  long long sum = 0;
  int cur_val = -1;
  long long result = 0;
  multiset<int> times;
  for (int i = 0; i < n;) {
    while (times.size() != 0 && arr[i].first != cur_val) {
      cur_val++;
      auto it = times.end();
      --it;
      sum -= *it;
      result += sum;
      times.erase(it);
    }
    if (times.size() == 0) {
      cur_val = arr[i].first;
    }
    times.insert(t[arr[i].second]);
    sum += t[arr[i].second];
    i++;
    while (i < n && arr[i].first == cur_val) {
      times.insert(t[arr[i].second]);
      sum += t[arr[i++].second];
    }
    cur_val++;
    auto it = times.end();
    --it;
    sum -= *it;
    result += sum;
    times.erase(it);
  }
  while (times.size() != 0) {
    cur_val++;
    auto it = times.end();
    --it;
    sum -= *it;
    result += sum;
    times.erase(it);
  }
  cout << result;
  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() {
  int n;
  cin >> n;
  vector<pair<int, int>> arr;
  arr.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i].first;
    arr[i].second = i;
  }
  vector<int> t;
  t.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  sort(arr.begin(), arr.end());
  long long sum = 0;
  int cur_val = -1;
  long long result = 0;
  multiset<int> times;
  for (int i = 0; i < n;) {
    while (times.size() != 0 && arr[i].first != cur_val) {
      cur_val++;
      auto it = times.end();
      --it;
      sum -= *it;
      result += sum;
      times.erase(it);
    }
    if (times.size() == 0) {
      cur_val = arr[i].first;
    }
    times.insert(t[arr[i].second]);
    sum += t[arr[i].second];
    i++;
    while (i < n && arr[i].first == cur_val) {
      times.insert(t[arr[i].second]);
      sum += t[arr[i++].second];
    }
    cur_val++;
    auto it = times.end();
    --it;
    sum -= *it;
    result += sum;
    times.erase(it);
  }
  while (times.size() != 0) {
    cur_val++;
    auto it = times.end();
    --it;
    sum -= *it;
    result += sum;
    times.erase(it);
  }
  cout << result;
  return 0;
}

```