#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int &x : arr) {
    cin >> x;
  }
  map<int, vector<int>> cost;
  for (int i = 0; i < n; i++) {
    int tmp;
    cin >> tmp;
    cost[arr[i]].push_back(tmp);
  }
  long long price = 0, curp = 0;
  sort(arr.begin(), arr.end());
  arr.erase(unique(arr.begin(), arr.end()), arr.end());
  priority_queue<long long> tmp;
  for (auto x : arr) {
    for (auto i = 0; i < cost[x].size(); i++) {
      curp += cost[x][i];
      tmp.push(cost[x][i]);
    }
    curp -= tmp.top();
    int dist = x;
    dist++;
    tmp.pop();
    int ext = 0;
    while (!tmp.empty()) {
      price += curp;
      if (cost[dist].size()) {
        break;
      } else {
        curp -= tmp.top();
        dist++;
        tmp.pop();
      }
    }
  }
  cout << price << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int testcase = 1;
  for (int i = 0; i < testcase; i++) {
    solve();
  }
  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;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int &x : arr) {
    cin >> x;
  }
  map<int, vector<int>> cost;
  for (int i = 0; i < n; i++) {
    int tmp;
    cin >> tmp;
    cost[arr[i]].push_back(tmp);
  }
  long long price = 0, curp = 0;
  sort(arr.begin(), arr.end());
  arr.erase(unique(arr.begin(), arr.end()), arr.end());
  priority_queue<long long> tmp;
  for (auto x : arr) {
    for (auto i = 0; i < cost[x].size(); i++) {
      curp += cost[x][i];
      tmp.push(cost[x][i]);
    }
    curp -= tmp.top();
    int dist = x;
    dist++;
    tmp.pop();
    int ext = 0;
    while (!tmp.empty()) {
      price += curp;
      if (cost[dist].size()) {
        break;
      } else {
        curp -= tmp.top();
        dist++;
        tmp.pop();
      }
    }
  }
  cout << price << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int testcase = 1;
  for (int i = 0; i < testcase; i++) {
    solve();
  }
  return 0;
}

```