#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:268435456")
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2>& p) {
  out << p.first << " " << p.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& v) {
  for (int i = 0; i < int(v.size()); ++i) {
    in >> v[i];
  }
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T>& v) {
  for (int i = 0; i < int(v.size()); ++i) {
    out << v[i] << ' ';
  }
  return out;
}
vector<int> a, t;
bool cmp(int a, int b) { return t[a] < t[b]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  a.resize(n);
  t.resize(n);
  cin >> a >> t;
  map<int, vector<int> > was;
  for (int i = 0; i < int(n); ++i) was[a[i]].push_back(i);
  long long res = 0;
  multiset<int> up;
  long long sum = 0;
  for (auto it = (was).begin(); it != (was).end(); ++it) {
    sort((it->second).begin(), (it->second).end(), cmp);
    if (!up.empty() && it->second.front() == -1) {
      sum -= *(--up.end());
      up.erase(--up.end());
    } else if (!up.empty() && t[it->second.back()] < *(--up.end())) {
      sum -= *(--up.end());
      up.erase(--up.end());
      for (int i = 0; i < int(it->second.size()); ++i) {
        up.insert(t[it->second[i]]);
        sum += t[it->second[i]];
      }
    } else {
      for (int i = 0; i < int(it->second.size() - 1); ++i) {
        up.insert(t[it->second[i]]);
        sum += t[it->second[i]];
      }
    }
    res += sum;
    auto it1 = it;
    ++it1;
    if (!up.empty() && (it1 == was.end() || it1->first - it->first > 1))
      was[it->first + 1].push_back(-1);
  }
  cout << res << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#pragma comment(linker, "/STACK:268435456")
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2>& p) {
  out << p.first << " " << p.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& v) {
  for (int i = 0; i < int(v.size()); ++i) {
    in >> v[i];
  }
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T>& v) {
  for (int i = 0; i < int(v.size()); ++i) {
    out << v[i] << ' ';
  }
  return out;
}
vector<int> a, t;
bool cmp(int a, int b) { return t[a] < t[b]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  a.resize(n);
  t.resize(n);
  cin >> a >> t;
  map<int, vector<int> > was;
  for (int i = 0; i < int(n); ++i) was[a[i]].push_back(i);
  long long res = 0;
  multiset<int> up;
  long long sum = 0;
  for (auto it = (was).begin(); it != (was).end(); ++it) {
    sort((it->second).begin(), (it->second).end(), cmp);
    if (!up.empty() && it->second.front() == -1) {
      sum -= *(--up.end());
      up.erase(--up.end());
    } else if (!up.empty() && t[it->second.back()] < *(--up.end())) {
      sum -= *(--up.end());
      up.erase(--up.end());
      for (int i = 0; i < int(it->second.size()); ++i) {
        up.insert(t[it->second[i]]);
        sum += t[it->second[i]];
      }
    } else {
      for (int i = 0; i < int(it->second.size() - 1); ++i) {
        up.insert(t[it->second[i]]);
        sum += t[it->second[i]];
      }
    }
    res += sum;
    auto it1 = it;
    ++it1;
    if (!up.empty() && (it1 == was.end() || it1->first - it->first > 1))
      was[it->first + 1].push_back(-1);
  }
  cout << res << endl;
  return 0;
}

```