#include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  solve();
}
void solve() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  map<long long, vector<long long>> mp;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    mp[v[i]].push_back(x);
  }
  long long ans = 0;
  priority_queue<long long> p;
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  long long res = 0;
  for (auto e : v) {
    if (mp[e].size() > 1 || !p.empty()) {
      for (int i = 0; i < mp[e].size(); ++i) {
        p.push(mp[e][i]);
        ans += mp[e][i];
      }
      ans -= p.top();
      p.pop();
      int curr = e;
      while (!p.empty()) {
        curr++;
        res += ans;
        if (mp[curr].size()) {
          break;
        } else {
          ans -= p.top();
          p.pop();
        }
      }
    }
  }
  cout << res;
}

### Prompt

Create a solution in CPP for 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();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  solve();
}
void solve() {
  int n;
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  map<long long, vector<long long>> mp;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    mp[v[i]].push_back(x);
  }
  long long ans = 0;
  priority_queue<long long> p;
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  long long res = 0;
  for (auto e : v) {
    if (mp[e].size() > 1 || !p.empty()) {
      for (int i = 0; i < mp[e].size(); ++i) {
        p.push(mp[e][i]);
        ans += mp[e][i];
      }
      ans -= p.top();
      p.pop();
      int curr = e;
      while (!p.empty()) {
        curr++;
        res += ans;
        if (mp[curr].size()) {
          break;
        } else {
          ans -= p.top();
          p.pop();
        }
      }
    }
  }
  cout << res;
}

```