#include <bits/stdc++.h>
using namespace std;
const double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1>
inline istream &operator>>(istream &in, pair<T0, T1> &a) {
  return in >> a.first >> a.second;
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < int(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(4) << fixed;
  smain();
  return 0;
}
const int N = 2e5 + 10;
map<long long, vector<long long> > ts;
void smain() {
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
  }
  vector<long long> t(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> t[i];
    ts[a[i]].push_back(t[i]);
  }
  multiset<long long> curs;
  long long sumt = 0;
  long long ans = 0;
  for (long long i = 1;; ++i, ans += sumt) {
    if (ts.count(i)) {
      for (auto t : ts[i]) {
        curs.insert(t);
        sumt += t;
      }
    }
    if (curs.empty()) {
      auto it = ts.lower_bound(i);
      if (it == ts.end()) break;
      i = (*it).first - 1;
      assert(sumt == 0);
    } else {
      auto it = prev(curs.end());
      sumt -= *it;
      curs.erase(it);
    }
  }
  cout << ans;
}

### 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;
const double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1>
inline istream &operator>>(istream &in, pair<T0, T1> &a) {
  return in >> a.first >> a.second;
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < int(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(4) << fixed;
  smain();
  return 0;
}
const int N = 2e5 + 10;
map<long long, vector<long long> > ts;
void smain() {
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
  }
  vector<long long> t(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> t[i];
    ts[a[i]].push_back(t[i]);
  }
  multiset<long long> curs;
  long long sumt = 0;
  long long ans = 0;
  for (long long i = 1;; ++i, ans += sumt) {
    if (ts.count(i)) {
      for (auto t : ts[i]) {
        curs.insert(t);
        sumt += t;
      }
    }
    if (curs.empty()) {
      auto it = ts.lower_bound(i);
      if (it == ts.end()) break;
      i = (*it).first - 1;
      assert(sumt == 0);
    } else {
      auto it = prev(curs.end());
      sumt -= *it;
      curs.erase(it);
    }
  }
  cout << ans;
}

```