#include <bits/stdc++.h>
using namespace std;
const int mx = 1e5 + 5;
class cmp {
 public:
  cmp() {}
  bool operator()(const pair<int, int>& p1, const pair<int, int>& p2) const {
    if (p1.first == p2.first)
      return p1.second > p2.second;
    else
      return (p1.first < p2.first);
  }
};
priority_queue<int> pq;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<pair<int, int>> v(n + 2);
  map<int, int> mp;
  for (int i = 1; i <= n; i++) {
    cin >> v[i].first;
  }
  for (int i = 1; i <= n; i++) {
    cin >> v[i].second;
  }
  v[0] = make_pair(0, 0);
  v[n + 1] = make_pair(2e9 + 100, 0);
  sort(v.begin(), v.end(), cmp());
  long long temp = 0;
  long long ans = 0;
  for (int i = 1; i <= n + 1; i++) {
    int f = v[i].first - v[i - 1].first;
    while (f && !pq.empty()) {
      temp -= pq.top();
      ans += temp;
      pq.pop();
      f--;
    }
    pq.push(v[i].second);
    temp += v[i].second;
  }
  cout << ans << endl;
  return 0;
}

### 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;
const int mx = 1e5 + 5;
class cmp {
 public:
  cmp() {}
  bool operator()(const pair<int, int>& p1, const pair<int, int>& p2) const {
    if (p1.first == p2.first)
      return p1.second > p2.second;
    else
      return (p1.first < p2.first);
  }
};
priority_queue<int> pq;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<pair<int, int>> v(n + 2);
  map<int, int> mp;
  for (int i = 1; i <= n; i++) {
    cin >> v[i].first;
  }
  for (int i = 1; i <= n; i++) {
    cin >> v[i].second;
  }
  v[0] = make_pair(0, 0);
  v[n + 1] = make_pair(2e9 + 100, 0);
  sort(v.begin(), v.end(), cmp());
  long long temp = 0;
  long long ans = 0;
  for (int i = 1; i <= n + 1; i++) {
    int f = v[i].first - v[i - 1].first;
    while (f && !pq.empty()) {
      temp -= pq.top();
      ans += temp;
      pq.pop();
      f--;
    }
    pq.push(v[i].second);
    temp += v[i].second;
  }
  cout << ans << endl;
  return 0;
}

```