#include <bits/stdc++.h>
using namespace std;
void solve() {
  int N;
  cin >> N;
  vector<int> A(N), T(N);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  for (int i = 0; i < N; i++) {
    cin >> T[i];
  }
  vector<int> ord(N);
  iota(ord.begin(), ord.end(), 0);
  sort(ord.begin(), ord.end(),
       [&](const int &i, const int &j) { return A[i] < A[j]; });
  multiset<int> Ts;
  int pnt = 0, mn;
  long long ans = 0;
  long long sum = 0;
  while (pnt < N || !Ts.empty()) {
    if (Ts.empty()) {
      mn = A[ord[pnt]];
    }
    while (pnt < N && A[ord[pnt]] == mn) {
      sum += T[ord[pnt]];
      Ts.insert(T[ord[pnt]]);
      pnt++;
    }
    sum -= *Ts.rbegin();
    Ts.erase(Ts.find(*Ts.rbegin()));
    ans += sum;
    mn++;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
void solve() {
  int N;
  cin >> N;
  vector<int> A(N), T(N);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  for (int i = 0; i < N; i++) {
    cin >> T[i];
  }
  vector<int> ord(N);
  iota(ord.begin(), ord.end(), 0);
  sort(ord.begin(), ord.end(),
       [&](const int &i, const int &j) { return A[i] < A[j]; });
  multiset<int> Ts;
  int pnt = 0, mn;
  long long ans = 0;
  long long sum = 0;
  while (pnt < N || !Ts.empty()) {
    if (Ts.empty()) {
      mn = A[ord[pnt]];
    }
    while (pnt < N && A[ord[pnt]] == mn) {
      sum += T[ord[pnt]];
      Ts.insert(T[ord[pnt]]);
      pnt++;
    }
    sum -= *Ts.rbegin();
    Ts.erase(Ts.find(*Ts.rbegin()));
    ans += sum;
    mn++;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

```