#include <bits/stdc++.h>
using namespace std;
using nagai = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<nagai, nagai>> a(n);
  vector<nagai> b(n);
  for (auto& x : a) cin >> x.first;
  for (auto& x : a) cin >> x.second;
  sort(a.begin(), a.end());
  multiset<nagai> nums;
  nagai ans = 0;
  map<nagai, vector<nagai>> v;
  for (auto& x : a) v[x.first].push_back(x.second);
  v[(int)1.5e9];
  int last = -1e9;
  nagai sm = 0;
  for (auto [x, y] : v) {
    for (int i = last; i < x; ++i) {
      if (nums.size()) {
        sm -= *prev(nums.end());
        nums.erase(prev(nums.end()));
      } else
        break;
      ans += sm;
    }
    for (auto x : y) nums.insert(x), sm += x;
    last = x;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
using nagai = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<nagai, nagai>> a(n);
  vector<nagai> b(n);
  for (auto& x : a) cin >> x.first;
  for (auto& x : a) cin >> x.second;
  sort(a.begin(), a.end());
  multiset<nagai> nums;
  nagai ans = 0;
  map<nagai, vector<nagai>> v;
  for (auto& x : a) v[x.first].push_back(x.second);
  v[(int)1.5e9];
  int last = -1e9;
  nagai sm = 0;
  for (auto [x, y] : v) {
    for (int i = last; i < x; ++i) {
      if (nums.size()) {
        sm -= *prev(nums.end());
        nums.erase(prev(nums.end()));
      } else
        break;
      ans += sm;
    }
    for (auto x : y) nums.insert(x), sm += x;
    last = x;
  }
  cout << ans << '\n';
  return 0;
}

```