#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  map<int, vector<int>> mp;
  for (int i = 0; i < n; i++) mp[a[i]].push_back(b[i]);
  priority_queue<int> que;
  long long ans = 0;
  long long sum = 0;
  int val = 0;
  for (auto& p : mp) {
    while (que.size() && val < p.first) {
      val++;
      ans += sum;
      sum -= que.top();
      que.pop();
    }
    val = p.first + 1;
    ans += sum;
    for (int x : p.second) {
      que.push(x);
      sum += x;
    }
    sum -= que.top();
    que.pop();
  }
  while (que.size()) {
    ans += sum;
    sum -= que.top();
    que.pop();
  }
  cout << ans << endl;
  return 0;
}

### 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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  map<int, vector<int>> mp;
  for (int i = 0; i < n; i++) mp[a[i]].push_back(b[i]);
  priority_queue<int> que;
  long long ans = 0;
  long long sum = 0;
  int val = 0;
  for (auto& p : mp) {
    while (que.size() && val < p.first) {
      val++;
      ans += sum;
      sum -= que.top();
      que.pop();
    }
    val = p.first + 1;
    ans += sum;
    for (int x : p.second) {
      que.push(x);
      sum += x;
    }
    sum -= que.top();
    que.pop();
  }
  while (que.size()) {
    ans += sum;
    sum -= que.top();
    que.pop();
  }
  cout << ans << endl;
  return 0;
}

```