#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<pair<long long, long long>> val(n);
  vector<long long> ai(n), ti(n);
  for (int i = 0; i < n; i++) cin >> ai[i];
  for (int i = 0; i < n; i++) cin >> ti[i];
  for (int i = 0; i < n; i++) val[i].first = ai[i], val[i].second = ti[i];
  sort(val.begin(), val.end());
  long long cur = 1;
  priority_queue<long long> pq;
  long long ans = 0, sm = 0;
  int idx = 0;
  while (idx < n || pq.empty() == false) {
    if (idx >= n) {
      long long tp = pq.top();
      pq.pop();
      sm = sm - tp;
      ans += sm;
      cur++;
      continue;
    }
    if (val[idx].first > cur && pq.empty() == false) {
      long long tp = pq.top();
      pq.pop();
      sm = sm - tp;
      ans += sm;
      cur++;
      continue;
    }
    if (val[idx].first > cur) {
      cur = val[idx].first;
    }
    while (idx < n && cur == val[idx].first) {
      pq.push(val[idx].second);
      sm += val[idx].second;
      idx++;
    }
    long long tp = pq.top();
    pq.pop();
    sm = sm - tp;
    ans += sm;
    cur++;
  }
  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;
int main() {
  int n;
  cin >> n;
  vector<pair<long long, long long>> val(n);
  vector<long long> ai(n), ti(n);
  for (int i = 0; i < n; i++) cin >> ai[i];
  for (int i = 0; i < n; i++) cin >> ti[i];
  for (int i = 0; i < n; i++) val[i].first = ai[i], val[i].second = ti[i];
  sort(val.begin(), val.end());
  long long cur = 1;
  priority_queue<long long> pq;
  long long ans = 0, sm = 0;
  int idx = 0;
  while (idx < n || pq.empty() == false) {
    if (idx >= n) {
      long long tp = pq.top();
      pq.pop();
      sm = sm - tp;
      ans += sm;
      cur++;
      continue;
    }
    if (val[idx].first > cur && pq.empty() == false) {
      long long tp = pq.top();
      pq.pop();
      sm = sm - tp;
      ans += sm;
      cur++;
      continue;
    }
    if (val[idx].first > cur) {
      cur = val[idx].first;
    }
    while (idx < n && cur == val[idx].first) {
      pq.push(val[idx].second);
      sm += val[idx].second;
      idx++;
    }
    long long tp = pq.top();
    pq.pop();
    sm = sm - tp;
    ans += sm;
    cur++;
  }
  cout << ans << endl;
  return 0;
}

```