#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  map<int, vector<int>> events;
  vector<pair<int, int>> a(n);
  int prev = 1e9;
  for (int i = (int)0; i < (int)n; i++) {
    cin >> a[i].first;
    prev = min(prev, a[i].first);
  }
  for (int i = (int)0; i < (int)n; i++) cin >> a[i].second;
  for (int i = (int)0; i < (int)n; i++)
    events[a[i].first].push_back(a[i].second);
  long long sum = 0, ans = 0;
  multiset<int> cur;
  for (auto& [idx, b] : events) {
    for (int i = (int)0; i < (int)idx - prev; i++) {
      if (cur.empty()) break;
      ans += 1LL * i * (*cur.rbegin());
      sum -= (*cur.rbegin());
      cur.erase(--cur.end());
    }
    ans += (idx - prev) * sum;
    for (int j : b) {
      cur.insert(j);
      sum += j;
    }
    prev = idx;
  }
  int i = 0;
  while (!cur.empty()) {
    ans += 1LL * i * (*cur.rbegin());
    sum -= (*cur.rbegin());
    cur.erase(--cur.end());
    ++i;
  }
  cout << ans;
  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>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  map<int, vector<int>> events;
  vector<pair<int, int>> a(n);
  int prev = 1e9;
  for (int i = (int)0; i < (int)n; i++) {
    cin >> a[i].first;
    prev = min(prev, a[i].first);
  }
  for (int i = (int)0; i < (int)n; i++) cin >> a[i].second;
  for (int i = (int)0; i < (int)n; i++)
    events[a[i].first].push_back(a[i].second);
  long long sum = 0, ans = 0;
  multiset<int> cur;
  for (auto& [idx, b] : events) {
    for (int i = (int)0; i < (int)idx - prev; i++) {
      if (cur.empty()) break;
      ans += 1LL * i * (*cur.rbegin());
      sum -= (*cur.rbegin());
      cur.erase(--cur.end());
    }
    ans += (idx - prev) * sum;
    for (int j : b) {
      cur.insert(j);
      sum += j;
    }
    prev = idx;
  }
  int i = 0;
  while (!cur.empty()) {
    ans += 1LL * i * (*cur.rbegin());
    sum -= (*cur.rbegin());
    cur.erase(--cur.end());
    ++i;
  }
  cout << ans;
  return 0;
}

```