#include <bits/stdc++.h>
using namespace std;
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n = nxt();
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; ++i) {
    a[i].first = nxt();
  }
  for (int i = 0; i < n; ++i) {
    a[i].second = nxt();
  }
  sort((a).begin(), (a).end());
  long long ans = 0;
  priority_queue<int> S;
  long long income = 0;
  for (int i = 0, cur = 0; i < n || !S.empty();) {
    if (i < n && S.empty() && cur < a[i].first) {
      cur = a[i].first;
    }
    while (i < n && a[i].first == cur) {
      S.push(a[i].second);
      income += a[i].second;
      ++i;
    }
    ++cur;
    income -= S.top();
    S.pop();
    ans += income;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please create a solution in CPP 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;
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n = nxt();
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; ++i) {
    a[i].first = nxt();
  }
  for (int i = 0; i < n; ++i) {
    a[i].second = nxt();
  }
  sort((a).begin(), (a).end());
  long long ans = 0;
  priority_queue<int> S;
  long long income = 0;
  for (int i = 0, cur = 0; i < n || !S.empty();) {
    if (i < n && S.empty() && cur < a[i].first) {
      cur = a[i].first;
    }
    while (i < n && a[i].first == cur) {
      S.push(a[i].second);
      income += a[i].second;
      ++i;
    }
    ++cur;
    income -= S.top();
    S.pop();
    ans += income;
  }
  cout << ans << "\n";
  return 0;
}

```