#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) {
    return a.second < b.second;
  }
  return a.first < b.first;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> b(n);
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
  }
  map<long long, vector<long long> > mp;
  for (long long i = 0; i < n; i++) {
    mp[a[i]].push_back(b[i]);
  }
  multiset<long long> old;
  long long sum = 0;
  long long res = 0;
  long long last = *min_element(a.begin(), a.end());
  for (auto x : mp) {
    if (x.first == last) {
      for (long long y : x.second) {
        old.insert(y);
        sum += y;
      }
      auto it = old.end();
      it--;
      sum -= (*it);
      old.erase(it);
      last = x.first;
      res += sum;
    } else {
      while (old.size() > 0 and last < x.first - 1) {
        auto it = old.end();
        it--;
        sum -= (*it);
        res += sum;
        old.erase(it);
        last++;
      }
      for (long long y : x.second) {
        old.insert(y);
        sum += y;
      }
      auto it = old.end();
      it--;
      sum -= (*it);
      old.erase(it);
      last = x.first;
      res += sum;
    }
  }
  while (old.size() > 0) {
    auto it = old.end();
    it--;
    sum -= (*it);
    res += sum;
    old.erase(it);
  }
  cout << res << '\n';
  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;
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) {
    return a.second < b.second;
  }
  return a.first < b.first;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> b(n);
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
  }
  map<long long, vector<long long> > mp;
  for (long long i = 0; i < n; i++) {
    mp[a[i]].push_back(b[i]);
  }
  multiset<long long> old;
  long long sum = 0;
  long long res = 0;
  long long last = *min_element(a.begin(), a.end());
  for (auto x : mp) {
    if (x.first == last) {
      for (long long y : x.second) {
        old.insert(y);
        sum += y;
      }
      auto it = old.end();
      it--;
      sum -= (*it);
      old.erase(it);
      last = x.first;
      res += sum;
    } else {
      while (old.size() > 0 and last < x.first - 1) {
        auto it = old.end();
        it--;
        sum -= (*it);
        res += sum;
        old.erase(it);
        last++;
      }
      for (long long y : x.second) {
        old.insert(y);
        sum += y;
      }
      auto it = old.end();
      it--;
      sum -= (*it);
      old.erase(it);
      last = x.first;
      res += sum;
    }
  }
  while (old.size() > 0) {
    auto it = old.end();
    it--;
    sum -= (*it);
    res += sum;
    old.erase(it);
  }
  cout << res << '\n';
  return 0;
}

```