#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long int, long long int> > v;
  unordered_map<long long, vector<long long int> > m;
  for (long long int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    v.push_back({x, 0});
  }
  for (long long int i = 0; i < n; i++) {
    cin >> v[i].second;
  }
  long long an = 0, an2 = 0;
  set<pair<long long int, long long int> > s;
  sort((v).begin(), (v).end());
  long long x = 0, y = 0;
  s.insert({v[0].second, 0});
  x = 1, y = v[0].first;
  an2 = v[0].second;
  while (!s.empty() || x < n) {
    while (x < n && y == v[x].first) {
      s.insert({v[x].second, x});
      an2 += v[x].second;
      x++;
    }
    y++;
    auto i = --s.end();
    an2 -= ((*s.rbegin()).first);
    an += an2;
    s.erase(i);
    if (s.empty() && x < n) y = v[x].first;
  }
  cout << an << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TESTS = 1, an = 1;
  while (TESTS--) {
    an++;
    solve();
  }
  return 0;
}

### 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;
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long int, long long int> > v;
  unordered_map<long long, vector<long long int> > m;
  for (long long int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    v.push_back({x, 0});
  }
  for (long long int i = 0; i < n; i++) {
    cin >> v[i].second;
  }
  long long an = 0, an2 = 0;
  set<pair<long long int, long long int> > s;
  sort((v).begin(), (v).end());
  long long x = 0, y = 0;
  s.insert({v[0].second, 0});
  x = 1, y = v[0].first;
  an2 = v[0].second;
  while (!s.empty() || x < n) {
    while (x < n && y == v[x].first) {
      s.insert({v[x].second, x});
      an2 += v[x].second;
      x++;
    }
    y++;
    auto i = --s.end();
    an2 -= ((*s.rbegin()).first);
    an += an2;
    s.erase(i);
    if (s.empty() && x < n) y = v[x].first;
  }
  cout << an << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TESTS = 1, an = 1;
  while (TESTS--) {
    an++;
    solve();
  }
  return 0;
}

```