#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, a[200005], t[200005];
  cin >> n;
  vector<pair<int, int>> ats;
  ats.reserve(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
    ats.push_back({a[i], t[i]});
  }
  sort(ats.begin(), ats.end(), [](pair<int, int> &a, pair<int, int> &b) {
    return a.first == b.first ? a.second > b.second : a.first < b.first;
  });
  long long res = 0;
  priority_queue<pair<int, int>> pq;
  for (int i = 0; i < n; ++i) {
    pq.push({ats[i].second, ats[i].first});
    int j = i + 1;
    long long l = ats[i].first, r;
    while (j < n && ats[j].first == l) {
      pq.push({ats[j].second, ats[j].first});
      ++j;
    }
    if (j >= n) {
      r = INT_MAX;
    } else {
      r = ats[j].first;
    }
    long long curA = l, curT;
    while (curA < r && !pq.empty()) {
      auto p = pq.top();
      pq.pop();
      curT = p.first;
      res += (curA - p.second) * curT;
      ++curA;
    }
    i = j - 1;
  }
  cout << res << endl;
  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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, a[200005], t[200005];
  cin >> n;
  vector<pair<int, int>> ats;
  ats.reserve(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
    ats.push_back({a[i], t[i]});
  }
  sort(ats.begin(), ats.end(), [](pair<int, int> &a, pair<int, int> &b) {
    return a.first == b.first ? a.second > b.second : a.first < b.first;
  });
  long long res = 0;
  priority_queue<pair<int, int>> pq;
  for (int i = 0; i < n; ++i) {
    pq.push({ats[i].second, ats[i].first});
    int j = i + 1;
    long long l = ats[i].first, r;
    while (j < n && ats[j].first == l) {
      pq.push({ats[j].second, ats[j].first});
      ++j;
    }
    if (j >= n) {
      r = INT_MAX;
    } else {
      r = ats[j].first;
    }
    long long curA = l, curT;
    while (curA < r && !pq.empty()) {
      auto p = pq.top();
      pq.pop();
      curT = p.first;
      res += (curA - p.second) * curT;
      ++curA;
    }
    i = j - 1;
  }
  cout << res << endl;
  return 0;
}

```