#include <bits/stdc++.h>
using namespace std;
void showpq(priority_queue<long long> gq) {
  priority_queue<long long> g = gq;
  while (!g.empty()) {
    cout << '\t' << g.top();
    g.pop();
  }
  cout << '\n';
}
int main() {
  long long n;
  cin >> n;
  pair<long long, long long> a[n];
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  priority_queue<long long> p;
  long long sum = 0;
  long long ans = 0;
  long long cur = 0;
  long long i = 0;
  sort(a, a + n);
  while (i < n) {
    while (p.size() && cur < a[i].first) {
      long long k = p.top();
      p.pop();
      sum = sum - k;
      ans = ans + sum;
      cur++;
    }
    if (cur < a[i].first) cur = a[i].first;
    long long j;
    for (j = i; j < n; j++) {
      if (a[j].first == a[i].first) {
        p.push(a[j].second);
        sum = sum + a[j].second;
      } else
        break;
    }
    i = j;
  }
  while (p.size()) {
    int k = p.top();
    p.pop();
    sum = sum - k;
    ans = ans + sum;
  }
  cout << ans << endl;
  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;
void showpq(priority_queue<long long> gq) {
  priority_queue<long long> g = gq;
  while (!g.empty()) {
    cout << '\t' << g.top();
    g.pop();
  }
  cout << '\n';
}
int main() {
  long long n;
  cin >> n;
  pair<long long, long long> a[n];
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  priority_queue<long long> p;
  long long sum = 0;
  long long ans = 0;
  long long cur = 0;
  long long i = 0;
  sort(a, a + n);
  while (i < n) {
    while (p.size() && cur < a[i].first) {
      long long k = p.top();
      p.pop();
      sum = sum - k;
      ans = ans + sum;
      cur++;
    }
    if (cur < a[i].first) cur = a[i].first;
    long long j;
    for (j = i; j < n; j++) {
      if (a[j].first == a[i].first) {
        p.push(a[j].second);
        sum = sum + a[j].second;
      } else
        break;
    }
    i = j;
  }
  while (p.size()) {
    int k = p.top();
    p.pop();
    sum = sum - k;
    ans = ans + sum;
  }
  cout << ans << endl;
  return 0;
}

```