#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 10;
long long arr[MAX], arr2[MAX];
int n;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  for (int i = 0; i < n; ++i) cin >> arr2[i];
  vector<pair<long long, long long> > vp;
  for (int i = 0; i < n; ++i) vp.push_back({arr[i], arr2[i]});
  sort(vp.begin(), vp.end());
  for (int i = 0; i < n; ++i) arr[i] = vp[i].first, arr2[i] = vp[i].second;
  int idx = 0;
  int cur = arr[0];
  long long ans = 0, sum = 0;
  priority_queue<long long> q;
  while (1) {
    ans += sum;
    while (idx < n && cur == arr[idx]) {
      sum += arr2[idx];
      q.push(arr2[idx]);
      idx++;
    }
    if (q.size() == 0) {
      if (idx == n) break;
      cur = arr[idx];
    } else {
      long long x = q.top();
      q.pop();
      sum -= x;
      cur++;
    }
  }
  return cout << ans << "\n", 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 MAX = 2e5 + 10;
long long arr[MAX], arr2[MAX];
int n;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  for (int i = 0; i < n; ++i) cin >> arr2[i];
  vector<pair<long long, long long> > vp;
  for (int i = 0; i < n; ++i) vp.push_back({arr[i], arr2[i]});
  sort(vp.begin(), vp.end());
  for (int i = 0; i < n; ++i) arr[i] = vp[i].first, arr2[i] = vp[i].second;
  int idx = 0;
  int cur = arr[0];
  long long ans = 0, sum = 0;
  priority_queue<long long> q;
  while (1) {
    ans += sum;
    while (idx < n && cur == arr[idx]) {
      sum += arr2[idx];
      q.push(arr2[idx]);
      idx++;
    }
    if (q.size() == 0) {
      if (idx == n) break;
      cur = arr[idx];
    } else {
      long long x = q.top();
      q.pop();
      sum -= x;
      cur++;
    }
  }
  return cout << ans << "\n", 0;
}

```