#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n), t(n);
  vector<pair<long long, long long> > vp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vp.push_back({a[i], i});
  }
  for (int i = 0; i < n; i++) cin >> t[i];
  long long ans = 0;
  sort(vp.begin(), vp.end());
  long long sum = 0;
  priority_queue<long long> pq;
  int i = 0;
  while (i < n) {
    int j;
    for (j = i; j < n; j++) {
      if (vp[i].first == vp[j].first) {
        sum += t[vp[j].second];
        pq.push(t[vp[j].second]);
      } else
        break;
    }
    long long num = vp[i].first;
    while (j < n && pq.empty() == 0 && num < vp[j].first) {
      sum -= pq.top();
      pq.pop();
      ans += sum;
      num++;
    }
    i = j;
  }
  while (pq.empty() == 0) {
    sum -= pq.top();
    pq.pop();
    ans += sum;
  }
  cout << ans << "\n";
}
int main(void) {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  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;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n), t(n);
  vector<pair<long long, long long> > vp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vp.push_back({a[i], i});
  }
  for (int i = 0; i < n; i++) cin >> t[i];
  long long ans = 0;
  sort(vp.begin(), vp.end());
  long long sum = 0;
  priority_queue<long long> pq;
  int i = 0;
  while (i < n) {
    int j;
    for (j = i; j < n; j++) {
      if (vp[i].first == vp[j].first) {
        sum += t[vp[j].second];
        pq.push(t[vp[j].second]);
      } else
        break;
    }
    long long num = vp[i].first;
    while (j < n && pq.empty() == 0 && num < vp[j].first) {
      sum -= pq.top();
      pq.pop();
      ans += sum;
      num++;
    }
    i = j;
  }
  while (pq.empty() == 0) {
    sum -= pq.top();
    pq.pop();
    ans += sum;
  }
  cout << ans << "\n";
}
int main(void) {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```