#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
const long long nax = 1e3 + 69;
const long long inf = 9e18 + 69;
using pii = pair<long long, long long>;
void solve() {
  long long n;
  cin >> n;
  vector<pii> a(n);
  map<long long, long long> m;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
    m[a[i].first]++;
  }
  for (int i = 0; i < n; ++i) {
    cin >> a[i].second;
  }
  sort(a.begin(), a.end());
  multiset<long long> s;
  long long sum = a[0].second, ans = 0, i = 1, val = a[0].first;
  s.insert(a[0].second);
  while (i < n) {
    if (val == a[i].first) {
      s.insert(a[i].second);
      sum += a[i].second;
      i++;
    } else {
      if (s.empty()) {
        val = a[i].first;
        continue;
      }
      auto it = s.end();
      --it;
      sum -= *it;
      s.erase(it);
      ans += sum;
      val++;
    }
  }
  while (s.size()) {
    auto it = s.end();
    --it;
    sum -= *it;
    ans += sum;
    s.erase(it);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1, t1 = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
const long long M = 1e9 + 7;
const long long nax = 1e3 + 69;
const long long inf = 9e18 + 69;
using pii = pair<long long, long long>;
void solve() {
  long long n;
  cin >> n;
  vector<pii> a(n);
  map<long long, long long> m;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
    m[a[i].first]++;
  }
  for (int i = 0; i < n; ++i) {
    cin >> a[i].second;
  }
  sort(a.begin(), a.end());
  multiset<long long> s;
  long long sum = a[0].second, ans = 0, i = 1, val = a[0].first;
  s.insert(a[0].second);
  while (i < n) {
    if (val == a[i].first) {
      s.insert(a[i].second);
      sum += a[i].second;
      i++;
    } else {
      if (s.empty()) {
        val = a[i].first;
        continue;
      }
      auto it = s.end();
      --it;
      sum -= *it;
      s.erase(it);
      ans += sum;
      val++;
    }
  }
  while (s.size()) {
    auto it = s.end();
    --it;
    sum -= *it;
    ans += sum;
    s.erase(it);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1, t1 = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```