#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using uint = unsigned int;
using ull = unsigned long long;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
default_random_engine generator;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(14) << fixed;
  cerr << setprecision(14) << fixed;
  int tests = 1;
  while (tests--) {
    solve();
  }
}
struct Data {
  int a, t;
};
bool operator<(const Data &a, const Data &b) { return a.t > b.t; }
void solve() {
  int n;
  cin >> n;
  vector<Data> d(n);
  for (int i = 0; i < n; ++i) {
    cin >> d[i].a;
  }
  for (int i = 0; i < n; ++i) {
    cin >> d[i].t;
  }
  sort((d).begin(), (d).end());
  set<int> pos;
  set<int> nxt;
  ll res = 0;
  for (auto [a, t] : d) {
    if (!pos.count(a)) {
      pos.insert(a);
      nxt.erase(a);
      if (!pos.count(a + 1)) {
        nxt.insert(a + 1);
      }
    } else {
      auto it = nxt.lower_bound(a);
      int b = *it;
      pos.insert(b);
      nxt.erase(b);
      res += ll(b - a) * t;
      if (!pos.count(b + 1)) {
        nxt.insert(b + 1);
      }
    }
  }
  cout << res << endl;
}

### Prompt

Generate 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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using uint = unsigned int;
using ull = unsigned long long;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
default_random_engine generator;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(14) << fixed;
  cerr << setprecision(14) << fixed;
  int tests = 1;
  while (tests--) {
    solve();
  }
}
struct Data {
  int a, t;
};
bool operator<(const Data &a, const Data &b) { return a.t > b.t; }
void solve() {
  int n;
  cin >> n;
  vector<Data> d(n);
  for (int i = 0; i < n; ++i) {
    cin >> d[i].a;
  }
  for (int i = 0; i < n; ++i) {
    cin >> d[i].t;
  }
  sort((d).begin(), (d).end());
  set<int> pos;
  set<int> nxt;
  ll res = 0;
  for (auto [a, t] : d) {
    if (!pos.count(a)) {
      pos.insert(a);
      nxt.erase(a);
      if (!pos.count(a + 1)) {
        nxt.insert(a + 1);
      }
    } else {
      auto it = nxt.lower_bound(a);
      int b = *it;
      pos.insert(b);
      nxt.erase(b);
      res += ll(b - a) * t;
      if (!pos.count(b + 1)) {
        nxt.insert(b + 1);
      }
    }
  }
  cout << res << endl;
}

```