#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  for (auto &x : a) cin >> x.first;
  for (auto &x : a) cin >> x.second;
  long long ans = 0;
  sort(a.begin(), a.end());
  priority_queue<long long> pq;
  long long sum = 0;
  for (long long i = 0; i < n; ++i) {
    if (!pq.empty()) {
      long long dif = a[i].first - a[i - 1].first - 1;
      long long j = 0;
      while (!pq.empty()) {
        if (j >= dif) break;
        long long top = pq.top();
        pq.pop();
        ans += top * j;
        sum -= top;
        ++j;
      }
      assert(sum >= 0);
      ans += sum * dif;
    }
    long long j = i;
    while (j < n and a[i].first == a[j].first) {
      pq.push(a[j].second);
      sum += a[j].second;
      ++j;
    }
    long long top = pq.top();
    pq.pop();
    sum -= top;
    assert(sum >= 0);
    ans += sum;
    i = j - 1;
  }
  long long j = 0;
  while (!pq.empty()) {
    ans += pq.top() * j;
    ++j;
    pq.pop();
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  for (long long i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  for (auto &x : a) cin >> x.first;
  for (auto &x : a) cin >> x.second;
  long long ans = 0;
  sort(a.begin(), a.end());
  priority_queue<long long> pq;
  long long sum = 0;
  for (long long i = 0; i < n; ++i) {
    if (!pq.empty()) {
      long long dif = a[i].first - a[i - 1].first - 1;
      long long j = 0;
      while (!pq.empty()) {
        if (j >= dif) break;
        long long top = pq.top();
        pq.pop();
        ans += top * j;
        sum -= top;
        ++j;
      }
      assert(sum >= 0);
      ans += sum * dif;
    }
    long long j = i;
    while (j < n and a[i].first == a[j].first) {
      pq.push(a[j].second);
      sum += a[j].second;
      ++j;
    }
    long long top = pq.top();
    pq.pop();
    sum -= top;
    assert(sum >= 0);
    ans += sum;
    i = j - 1;
  }
  long long j = 0;
  while (!pq.empty()) {
    ans += pq.top() * j;
    ++j;
    pq.pop();
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  for (long long i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}

```