#include <bits/stdc++.h>
using namespace std;
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TESTS = 1;
  while (TESTS--) {
    long long n;
    cin >> n;
    vector<pair<long long, long long> > v(n);
    for (long long i = 0; i < n; i++) cin >> v[i].first;
    for (long long i = 0; i < n; i++) cin >> v[i].second;
    sort((v).begin(), (v).end());
    bool ok = 0;
    long long ans = 0;
    priority_queue<long long> pq;
    long long i = 0;
    long long sum = 0;
    while (1) {
      if (i == n - 1 and !ok) break;
      if (i >= n) break;
      if (v[i].first != v[i + 1].first and !ok) {
        i++;
        continue;
      }
      long long ind = -1;
      for (long long j = i; j < n; j++) {
        if (v[j].first == v[i].first) {
          pq.push(v[j].second);
          sum += v[j].second;
        } else {
          ind = j;
          break;
        }
      }
      if (ind == -1)
        i = n;
      else
        i = ind;
      if (ind == -1) {
        while (!pq.empty()) {
          sum -= pq.top();
          pq.pop();
          ans += sum;
        }
        break;
      }
      long long gap = v[ind].first - v[ind - 1].first;
      while (gap > 0 and !pq.empty()) {
        sum -= pq.top();
        pq.pop();
        ans += sum;
        gap--;
      }
      if (!pq.empty()) ok = 1;
    }
    cout << ans;
  }
}

### 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;
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TESTS = 1;
  while (TESTS--) {
    long long n;
    cin >> n;
    vector<pair<long long, long long> > v(n);
    for (long long i = 0; i < n; i++) cin >> v[i].first;
    for (long long i = 0; i < n; i++) cin >> v[i].second;
    sort((v).begin(), (v).end());
    bool ok = 0;
    long long ans = 0;
    priority_queue<long long> pq;
    long long i = 0;
    long long sum = 0;
    while (1) {
      if (i == n - 1 and !ok) break;
      if (i >= n) break;
      if (v[i].first != v[i + 1].first and !ok) {
        i++;
        continue;
      }
      long long ind = -1;
      for (long long j = i; j < n; j++) {
        if (v[j].first == v[i].first) {
          pq.push(v[j].second);
          sum += v[j].second;
        } else {
          ind = j;
          break;
        }
      }
      if (ind == -1)
        i = n;
      else
        i = ind;
      if (ind == -1) {
        while (!pq.empty()) {
          sum -= pq.top();
          pq.pop();
          ans += sum;
        }
        break;
      }
      long long gap = v[ind].first - v[ind - 1].first;
      while (gap > 0 and !pq.empty()) {
        sum -= pq.top();
        pq.pop();
        ans += sum;
        gap--;
      }
      if (!pq.empty()) ok = 1;
    }
    cout << ans;
  }
}

```