#include <bits/stdc++.h>
using namespace std;
const long long INFLL = 1e18;
const int MAXN = 1e6 + 100;
const long long INF = 1e9;
const long long mod1 = 1e9 + 7;
const long long mod2 = 2e9 + 11;
int main() {
  cin.tie(0);
  multiset<long long> s;
  long long sum = 0;
  int n;
  cin >> n;
  vector<pair<long long, long long> > a(n), t(n);
  for (int i = 0; i < (n); i++) cin >> a[i].first;
  for (int i = 0; i < (n); i++) cin >> a[i].second;
  sort((a).begin(), (a).end());
  vector<pair<long long, vector<long long> > > b;
  int i = 0;
  long long ans = 0;
  while (i < n) {
    vector<long long> c;
    int j = i;
    while (j < n && a[i].first == a[j].first) c.push_back(a[j].second), j++;
    b.push_back({a[i].first, c});
    i = j;
  }
  i = 0;
  while (i < b.size()) {
    if (i == 0) {
      for (auto to : b[i].second) s.insert(to), sum += to;
      ans += sum - (*s.rbegin());
      sum -= (*s.rbegin());
      auto to = s.end();
      to = prev(to);
      s.erase(to);
      long long mx = s.size();
      if (i < (int)(b.size()) - 1)
        mx = min(mx, b[i + 1].first - b[i].first - 1);
      for (int j = 0; j < mx; j++) {
        ans += sum - (*s.rbegin());
        sum -= *s.rbegin();
        auto to = s.end();
        to = prev(to);
        s.erase(to);
      }
    } else {
      for (auto to : b[i].second) s.insert(to), sum += to;
      ans += sum - (*s.rbegin());
      sum -= (*s.rbegin());
      auto to = s.end();
      to = prev(to);
      s.erase(to);
      long long mx = s.size();
      if (i < (int)(b.size()) - 1)
        mx = min(mx, b[i + 1].first - b[i].first - 1);
      for (int j = 0; j < mx; j++) {
        ans += sum - (*s.rbegin());
        sum -= *s.rbegin();
        auto to = s.end();
        to = prev(to);
        s.erase(to);
      }
    }
    i++;
  }
  cout << ans;
}

### 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;
const long long INFLL = 1e18;
const int MAXN = 1e6 + 100;
const long long INF = 1e9;
const long long mod1 = 1e9 + 7;
const long long mod2 = 2e9 + 11;
int main() {
  cin.tie(0);
  multiset<long long> s;
  long long sum = 0;
  int n;
  cin >> n;
  vector<pair<long long, long long> > a(n), t(n);
  for (int i = 0; i < (n); i++) cin >> a[i].first;
  for (int i = 0; i < (n); i++) cin >> a[i].second;
  sort((a).begin(), (a).end());
  vector<pair<long long, vector<long long> > > b;
  int i = 0;
  long long ans = 0;
  while (i < n) {
    vector<long long> c;
    int j = i;
    while (j < n && a[i].first == a[j].first) c.push_back(a[j].second), j++;
    b.push_back({a[i].first, c});
    i = j;
  }
  i = 0;
  while (i < b.size()) {
    if (i == 0) {
      for (auto to : b[i].second) s.insert(to), sum += to;
      ans += sum - (*s.rbegin());
      sum -= (*s.rbegin());
      auto to = s.end();
      to = prev(to);
      s.erase(to);
      long long mx = s.size();
      if (i < (int)(b.size()) - 1)
        mx = min(mx, b[i + 1].first - b[i].first - 1);
      for (int j = 0; j < mx; j++) {
        ans += sum - (*s.rbegin());
        sum -= *s.rbegin();
        auto to = s.end();
        to = prev(to);
        s.erase(to);
      }
    } else {
      for (auto to : b[i].second) s.insert(to), sum += to;
      ans += sum - (*s.rbegin());
      sum -= (*s.rbegin());
      auto to = s.end();
      to = prev(to);
      s.erase(to);
      long long mx = s.size();
      if (i < (int)(b.size()) - 1)
        mx = min(mx, b[i + 1].first - b[i].first - 1);
      for (int j = 0; j < mx; j++) {
        ans += sum - (*s.rbegin());
        sum -= *s.rbegin();
        auto to = s.end();
        to = prev(to);
        s.erase(to);
      }
    }
    i++;
  }
  cout << ans;
}

```