#include <bits/stdc++.h>
using namespace std;
const long long mxn = 1e5;
const long long INF = 2e18;
const int32_t M = 1000000007;
const long double pie = acos(-1);
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > v(n);
  for (long long i = 0; i <= n - 1; i++) {
    cin >> v[i].first;
  }
  for (long long i = 0; i <= n - 1; i++) {
    cin >> v[i].second;
  }
  sort((v).begin(), (v).end());
  long long mn = v[0].first;
  multiset<long long> t;
  long long ind = 0;
  long long ans = 0;
  long long sum = 0;
  while (ind < n) {
    while (ind < n && v[ind].first == mn) {
      t.insert(v[ind].second);
      sum += v[ind].second;
      ind++;
    }
    while (!t.empty()) {
      auto it = t.end();
      it--;
      sum -= (*it);
      ans += sum;
      t.erase(it);
      mn++;
      while (ind < n && v[ind].first == mn) {
        t.insert(v[ind].second);
        sum += v[ind].second;
        ind++;
      }
    }
    if (ind == n) break;
    mn = v[ind].first;
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long _t = 1;
  for (long long i = 1; i <= _t; i++) {
    solve();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 mxn = 1e5;
const long long INF = 2e18;
const int32_t M = 1000000007;
const long double pie = acos(-1);
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > v(n);
  for (long long i = 0; i <= n - 1; i++) {
    cin >> v[i].first;
  }
  for (long long i = 0; i <= n - 1; i++) {
    cin >> v[i].second;
  }
  sort((v).begin(), (v).end());
  long long mn = v[0].first;
  multiset<long long> t;
  long long ind = 0;
  long long ans = 0;
  long long sum = 0;
  while (ind < n) {
    while (ind < n && v[ind].first == mn) {
      t.insert(v[ind].second);
      sum += v[ind].second;
      ind++;
    }
    while (!t.empty()) {
      auto it = t.end();
      it--;
      sum -= (*it);
      ans += sum;
      t.erase(it);
      mn++;
      while (ind < n && v[ind].first == mn) {
        t.insert(v[ind].second);
        sum += v[ind].second;
        ind++;
      }
    }
    if (ind == n) break;
    mn = v[ind].first;
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long _t = 1;
  for (long long i = 1; i <= _t; i++) {
    solve();
  }
  return 0;
}

```