#include <bits/stdc++.h>
using namespace std;
int n, k;
const int maxn = 2e6 + 10;
long long book[maxn];
long long inter[maxn];
pair<long long, long long> interest[maxn];
bool cmp(const long long a, const long long b) { return book[a] < book[b]; }
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> book[i];
  for (int i = 0; i < n; ++i) cin >> inter[i];
  for (int i = 0; i < n; ++i) {
    interest[i] = make_pair(book[i], inter[i]);
  }
  sort(interest, interest + n);
  priority_queue<long long> pq;
  while (!pq.empty()) pq.pop();
  sort(book, book + n);
  pq.push(interest[0].second);
  long long ans = 0;
  long long sum = 0;
  sum += interest[0].second;
  for (long long i = 1; i < n; ++i) {
    if (book[i] != book[i - 1]) {
      if (!pq.empty()) {
        sum -= pq.top();
        pq.pop();
      }
      for (long long j = 0; j < book[i] - book[i - 1]; ++j) {
        if (!pq.empty()) {
          long long t = pq.top();
          ans += sum;
          if (book[i - 1] + j + 1 != book[i]) {
            sum -= t;
            pq.pop();
          }
        } else {
          break;
        }
      }
    }
    pq.push(interest[i].second);
    sum += interest[i].second;
  }
  if (!pq.empty()) {
    sum -= pq.top();
    pq.pop();
  }
  while (!pq.empty()) {
    long long t = pq.top();
    ans += sum;
    sum -= t;
    pq.pop();
  }
  cout << ans << "\n";
  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;
int n, k;
const int maxn = 2e6 + 10;
long long book[maxn];
long long inter[maxn];
pair<long long, long long> interest[maxn];
bool cmp(const long long a, const long long b) { return book[a] < book[b]; }
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> book[i];
  for (int i = 0; i < n; ++i) cin >> inter[i];
  for (int i = 0; i < n; ++i) {
    interest[i] = make_pair(book[i], inter[i]);
  }
  sort(interest, interest + n);
  priority_queue<long long> pq;
  while (!pq.empty()) pq.pop();
  sort(book, book + n);
  pq.push(interest[0].second);
  long long ans = 0;
  long long sum = 0;
  sum += interest[0].second;
  for (long long i = 1; i < n; ++i) {
    if (book[i] != book[i - 1]) {
      if (!pq.empty()) {
        sum -= pq.top();
        pq.pop();
      }
      for (long long j = 0; j < book[i] - book[i - 1]; ++j) {
        if (!pq.empty()) {
          long long t = pq.top();
          ans += sum;
          if (book[i - 1] + j + 1 != book[i]) {
            sum -= t;
            pq.pop();
          }
        } else {
          break;
        }
      }
    }
    pq.push(interest[i].second);
    sum += interest[i].second;
  }
  if (!pq.empty()) {
    sum -= pq.top();
    pq.pop();
  }
  while (!pq.empty()) {
    long long t = pq.top();
    ans += sum;
    sum -= t;
    pq.pop();
  }
  cout << ans << "\n";
  return 0;
}

```