#include <bits/stdc++.h>
using namespace std;
int n;
vector<pair<long long, long long> > v;
int j;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n;
  v.resize(n);
  for (int i = 0; i < n; i++) cin >> v[i].first;
  for (int i = 0; i < n; i++) cin >> v[i].second;
  sort(v.begin(), v.end());
  multiset<long long> s;
  long long ss = 0LL;
  long long res = 0LL;
  for (int i = 0; i < v.size();) {
    if (i == 0) {
      s.insert(v[i].second);
      ss += v[i].second;
      i++;
      continue;
    }
    if (v[i].first != v[i - 1].first) {
      long long vr = v[i - 1].first;
      while (vr < v[i].first && s.size() > 0) {
        multiset<long long>::iterator it;
        it = prev(s.end());
        ss -= *it;
        s.erase(it);
        res = res + ss;
        vr++;
      }
      if (vr == v[i].first) {
        for (j = i; j < v.size() && v[j].first == vr; j++) {
          s.insert(v[j].second);
          ss += v[j].second;
        }
        i = j;
      } else {
        s.insert(v[i].second);
        ss += v[i].second;
        i++;
      }
    } else {
      s.insert(v[i].second);
      ss += v[i].second;
      i++;
    }
  }
  while (s.size() > 0) {
    multiset<long long>::iterator it;
    it = prev(s.end());
    ss -= *it;
    s.erase(it);
    res = res + ss;
  }
  cout << res;
  return 0;
}

### Prompt

Please formulate 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;
int n;
vector<pair<long long, long long> > v;
int j;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n;
  v.resize(n);
  for (int i = 0; i < n; i++) cin >> v[i].first;
  for (int i = 0; i < n; i++) cin >> v[i].second;
  sort(v.begin(), v.end());
  multiset<long long> s;
  long long ss = 0LL;
  long long res = 0LL;
  for (int i = 0; i < v.size();) {
    if (i == 0) {
      s.insert(v[i].second);
      ss += v[i].second;
      i++;
      continue;
    }
    if (v[i].first != v[i - 1].first) {
      long long vr = v[i - 1].first;
      while (vr < v[i].first && s.size() > 0) {
        multiset<long long>::iterator it;
        it = prev(s.end());
        ss -= *it;
        s.erase(it);
        res = res + ss;
        vr++;
      }
      if (vr == v[i].first) {
        for (j = i; j < v.size() && v[j].first == vr; j++) {
          s.insert(v[j].second);
          ss += v[j].second;
        }
        i = j;
      } else {
        s.insert(v[i].second);
        ss += v[i].second;
        i++;
      }
    } else {
      s.insert(v[i].second);
      ss += v[i].second;
      i++;
    }
  }
  while (s.size() > 0) {
    multiset<long long>::iterator it;
    it = prev(s.end());
    ss -= *it;
    s.erase(it);
    res = res + ss;
  }
  cout << res;
  return 0;
}

```