#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
  }
  for (int i = 0; i < n; ++i) {
    cin >> a[i].second;
  }
  sort(a.begin(), a.end());
  long long ans = 0;
  multiset<int> p;
  int cV = -1;
  int nxtI = 0;
  long long pSum = 0;
  while (nxtI != n || !p.empty()) {
    if (p.empty()) {
      p.insert(-a[nxtI].second);
      pSum += a[nxtI].second;
      cV = a[nxtI].first;
      ++nxtI;
      while (nxtI < n && a[nxtI].first == cV) {
        p.insert(-a[nxtI].second);
        pSum += a[nxtI].second;
        ++nxtI;
      }
    } else {
      while (nxtI < n && a[nxtI].first == cV) {
        p.insert(-a[nxtI].second);
        pSum += a[nxtI].second;
        ++nxtI;
      }
      int x = -(*p.begin());
      auto ff = p.begin();
      ++ff;
      p.erase(p.begin(), ff);
      pSum -= x;
      ans += pSum;
      cV += 1;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
  }
  for (int i = 0; i < n; ++i) {
    cin >> a[i].second;
  }
  sort(a.begin(), a.end());
  long long ans = 0;
  multiset<int> p;
  int cV = -1;
  int nxtI = 0;
  long long pSum = 0;
  while (nxtI != n || !p.empty()) {
    if (p.empty()) {
      p.insert(-a[nxtI].second);
      pSum += a[nxtI].second;
      cV = a[nxtI].first;
      ++nxtI;
      while (nxtI < n && a[nxtI].first == cV) {
        p.insert(-a[nxtI].second);
        pSum += a[nxtI].second;
        ++nxtI;
      }
    } else {
      while (nxtI < n && a[nxtI].first == cV) {
        p.insert(-a[nxtI].second);
        pSum += a[nxtI].second;
        ++nxtI;
      }
      int x = -(*p.begin());
      auto ff = p.begin();
      ++ff;
      p.erase(p.begin(), ff);
      pSum -= x;
      ans += pSum;
      cV += 1;
    }
  }
  cout << ans << endl;
  return 0;
}

```