#include <bits/stdc++.h>
using namespace std;
const long long MX = 3e5;
struct Publication {
  long long a, t;
  bool operator<(const Publication &x) const { return t < x.t; }
} P[MX];
bool cmp(Publication &a, Publication &b) { return a.a < b.a; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 0; i < (long long)n; i++) {
    cin >> P[i].a;
  }
  for (long long i = 0; i < (long long)n; i++) {
    cin >> P[i].t;
  }
  sort(P, P + n, cmp);
  priority_queue<Publication> pq;
  long long sum = 0;
  long long i = 0;
  long long ans = 0;
  long long curA = -1;
  while (i < n || !pq.empty()) {
    if (pq.empty()) {
      long long a = P[i].a;
      curA = a;
      while (i < n && P[i].a == a) {
        sum += P[i].t;
        pq.push(P[i]);
        i++;
      }
    } else {
      if (i < n) {
        while (i < n && P[i].a == curA) {
          sum += P[i].t;
          pq.push(P[i]);
          i++;
        }
      }
      curA++;
      sum -= pq.top().t;
      pq.pop();
      ans += sum;
    }
  }
  cout << ans << 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>
using namespace std;
const long long MX = 3e5;
struct Publication {
  long long a, t;
  bool operator<(const Publication &x) const { return t < x.t; }
} P[MX];
bool cmp(Publication &a, Publication &b) { return a.a < b.a; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 0; i < (long long)n; i++) {
    cin >> P[i].a;
  }
  for (long long i = 0; i < (long long)n; i++) {
    cin >> P[i].t;
  }
  sort(P, P + n, cmp);
  priority_queue<Publication> pq;
  long long sum = 0;
  long long i = 0;
  long long ans = 0;
  long long curA = -1;
  while (i < n || !pq.empty()) {
    if (pq.empty()) {
      long long a = P[i].a;
      curA = a;
      while (i < n && P[i].a == a) {
        sum += P[i].t;
        pq.push(P[i]);
        i++;
      }
    } else {
      if (i < n) {
        while (i < n && P[i].a == curA) {
          sum += P[i].t;
          pq.push(P[i]);
          i++;
        }
      }
      curA++;
      sum -= pq.top().t;
      pq.pop();
      ans += sum;
    }
  }
  cout << ans << endl;
  return 0;
}

```