#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  long long int n;
  cin >> n;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int brr[n];
  for (long long int i = 0; i < n; i++) cin >> brr[i];
  map<long long int, vector<long long int> > collect;
  long long int answer = 0;
  set<long long int> s1;
  priority_queue<long long int> q1;
  for (long long int i = 0; i < n; i++) {
    s1.insert(arr[i]);
    collect[arr[i]].push_back(brr[i]);
  }
  long long int runval = -1;
  long long int sum = 0;
  for (auto itr = s1.begin(); itr != s1.end(); itr++) {
    long long int x = *itr;
    if (runval == -1) runval = x;
    while (q1.size() > 0 && runval < x) {
      sum -= q1.top();
      answer += sum;
      q1.pop();
      runval++;
    }
    runval = x;
    for (long long int j = 0; j < collect[x].size(); j++) {
      q1.push(collect[x][j]);
      sum += collect[x][j];
    }
    sum -= q1.top();
    q1.pop();
    runval += 1;
    answer += sum;
  }
  while (q1.size() > 0) {
    sum -= q1.top();
    answer += sum;
    q1.pop();
  }
  cout << answer << endl;
}

### Prompt

Your task is to create 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;
int32_t main() {
  long long int n;
  cin >> n;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int brr[n];
  for (long long int i = 0; i < n; i++) cin >> brr[i];
  map<long long int, vector<long long int> > collect;
  long long int answer = 0;
  set<long long int> s1;
  priority_queue<long long int> q1;
  for (long long int i = 0; i < n; i++) {
    s1.insert(arr[i]);
    collect[arr[i]].push_back(brr[i]);
  }
  long long int runval = -1;
  long long int sum = 0;
  for (auto itr = s1.begin(); itr != s1.end(); itr++) {
    long long int x = *itr;
    if (runval == -1) runval = x;
    while (q1.size() > 0 && runval < x) {
      sum -= q1.top();
      answer += sum;
      q1.pop();
      runval++;
    }
    runval = x;
    for (long long int j = 0; j < collect[x].size(); j++) {
      q1.push(collect[x][j]);
      sum += collect[x][j];
    }
    sum -= q1.top();
    q1.pop();
    runval += 1;
    answer += sum;
  }
  while (q1.size() > 0) {
    sum -= q1.top();
    answer += sum;
    q1.pop();
  }
  cout << answer << endl;
}

```