#include <bits/stdc++.h>
using namespace std;
const int INF = 2e9 + 239;
const long long INFL = 2e18 + 239;
signed main() {
  ios_base::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 ans = 0;
  multiset<long long> bag;
  long long sum = 0;
  for (long long j = 0; j < n;) {
    long long i = j;
    bag.insert(a[i].second);
    sum += a[i].second;
    while (i + 1 < n && a[i].first == a[i + 1].first) {
      bag.insert(a[i + 1].second);
      sum += a[i + 1].second;
      ++i;
    }
    long long num = a[i].first;
    while (i + 1 == n || num < a[i + 1].first) {
      if (bag.size() <= 1) {
        bag.clear();
        sum = 0;
        break;
      }
      sum -= *bag.rbegin();
      ans += sum;
      bag.erase(--bag.end());
      ++num;
    }
    j = i + 1;
  }
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 int INF = 2e9 + 239;
const long long INFL = 2e18 + 239;
signed main() {
  ios_base::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 ans = 0;
  multiset<long long> bag;
  long long sum = 0;
  for (long long j = 0; j < n;) {
    long long i = j;
    bag.insert(a[i].second);
    sum += a[i].second;
    while (i + 1 < n && a[i].first == a[i + 1].first) {
      bag.insert(a[i + 1].second);
      sum += a[i + 1].second;
      ++i;
    }
    long long num = a[i].first;
    while (i + 1 == n || num < a[i + 1].first) {
      if (bag.size() <= 1) {
        bag.clear();
        sum = 0;
        break;
      }
      sum -= *bag.rbegin();
      ans += sum;
      bag.erase(--bag.end());
      ++num;
    }
    j = i + 1;
  }
  cout << ans;
  return 0;
}

```