#include <bits/stdc++.h>
using namespace std;
multiset<long long> cost;
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  for (long long i = 0; i < n; i++) cin >> a[i].first;
  for (long long i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long res = 0;
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    long long cur_Ppl = a[i].first;
    while (i < n && cur_Ppl == a[i].first) {
      sum += a[i].second;
      cost.insert(a[i].second);
      i++;
    }
    long long last = *cost.rbegin();
    cost.erase(cost.find(last));
    sum -= last;
    long long sz = (long long)(cost.size());
    if (!sz) {
      i--;
      continue;
    }
    res += sum;
    if (i < n) {
      long long nxtPpl = a[i].first;
      while (sz && ++cur_Ppl < nxtPpl) {
        last = *cost.rbegin();
        cost.erase(cost.find(last));
        sum -= last;
        res += sum;
        sz--;
      }
    } else {
      while (sz) {
        last = *cost.rbegin();
        cost.erase(cost.find(last));
        sum -= last;
        res += sum;
        sz--;
      }
    }
    i--;
  }
  cout << res;
}

### Prompt

Your task is to create 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;
multiset<long long> cost;
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  for (long long i = 0; i < n; i++) cin >> a[i].first;
  for (long long i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long res = 0;
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    long long cur_Ppl = a[i].first;
    while (i < n && cur_Ppl == a[i].first) {
      sum += a[i].second;
      cost.insert(a[i].second);
      i++;
    }
    long long last = *cost.rbegin();
    cost.erase(cost.find(last));
    sum -= last;
    long long sz = (long long)(cost.size());
    if (!sz) {
      i--;
      continue;
    }
    res += sum;
    if (i < n) {
      long long nxtPpl = a[i].first;
      while (sz && ++cur_Ppl < nxtPpl) {
        last = *cost.rbegin();
        cost.erase(cost.find(last));
        sum -= last;
        res += sum;
        sz--;
      }
    } else {
      while (sz) {
        last = *cost.rbegin();
        cost.erase(cost.find(last));
        sum -= last;
        res += sum;
        sz--;
      }
    }
    i--;
  }
  cout << res;
}

```