#include <bits/stdc++.h>
using namespace std;
int main(void) {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long n, i = 0, j, k, x, y, z, ans = 0, sum = 0;
  bool flag = 0;
  vector<pair<long long, long long> > v;
  priority_queue<long long> q;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    v.push_back(make_pair(x, 1));
  }
  for (i = 0; i < n; i++) {
    cin >> x;
    v[i].second = x;
  }
  sort(v.begin(), v.end());
  v.push_back(make_pair(LLONG_MAX, LLONG_MAX));
  x = v[0].first;
  i = 0;
  while (i < n) {
    if (x == v[i].first) {
      q.push(v[i].second);
      sum += v[i].second;
      i++;
    } else {
      while (!q.empty() && x < v[i].first) {
        sum -= q.top();
        q.pop();
        ans += sum;
        x++;
      }
      x = v[i].first;
    }
  }
  while (!q.empty()) {
    sum -= q.top();
    q.pop();
    ans += sum;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate a CPP solution to 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(void) {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long n, i = 0, j, k, x, y, z, ans = 0, sum = 0;
  bool flag = 0;
  vector<pair<long long, long long> > v;
  priority_queue<long long> q;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    v.push_back(make_pair(x, 1));
  }
  for (i = 0; i < n; i++) {
    cin >> x;
    v[i].second = x;
  }
  sort(v.begin(), v.end());
  v.push_back(make_pair(LLONG_MAX, LLONG_MAX));
  x = v[0].first;
  i = 0;
  while (i < n) {
    if (x == v[i].first) {
      q.push(v[i].second);
      sum += v[i].second;
      i++;
    } else {
      while (!q.empty() && x < v[i].first) {
        sum -= q.top();
        q.pop();
        ans += sum;
        x++;
      }
      x = v[i].first;
    }
  }
  while (!q.empty()) {
    sum -= q.top();
    q.pop();
    ans += sum;
  }
  cout << ans << "\n";
  return 0;
}

```