#include <bits/stdc++.h>
using namespace std;
const long long inf = 4e18;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long a[n], i, j, o = 0;
  for (i = 0; i < n; i++) cin >> a[i];
  vector<pair<long long, long long> > v;
  for (i = 0; i < n; i++) {
    long long t;
    cin >> t;
    v.emplace_back(a[i], t);
  }
  multiset<long long, greater<long long> > ms;
  sort(v.begin(), v.end());
  ms.emplace(v[0].second);
  long long cnt = v[0].second;
  for (i = 1; i < n; i++) {
    if (v[i].first != v[i - 1].first) {
      long long c = v[i].first - v[i - 1].first;
      for (j = 0; j < c; j++) {
        o += j * *ms.begin();
        cnt -= *ms.begin();
        ms.erase(ms.begin());
        if (ms.size() == 0) break;
      }
      o += cnt * j;
    }
    ms.emplace(v[i].second);
    cnt += v[i].second;
  }
  for (j = 0; ms.size() > 0; j++) {
    o += j * *ms.begin();
    ms.erase(ms.begin());
  }
  cout << o << '\n';
}

### 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;
const long long inf = 4e18;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long a[n], i, j, o = 0;
  for (i = 0; i < n; i++) cin >> a[i];
  vector<pair<long long, long long> > v;
  for (i = 0; i < n; i++) {
    long long t;
    cin >> t;
    v.emplace_back(a[i], t);
  }
  multiset<long long, greater<long long> > ms;
  sort(v.begin(), v.end());
  ms.emplace(v[0].second);
  long long cnt = v[0].second;
  for (i = 1; i < n; i++) {
    if (v[i].first != v[i - 1].first) {
      long long c = v[i].first - v[i - 1].first;
      for (j = 0; j < c; j++) {
        o += j * *ms.begin();
        cnt -= *ms.begin();
        ms.erase(ms.begin());
        if (ms.size() == 0) break;
      }
      o += cnt * j;
    }
    ms.emplace(v[i].second);
    cnt += v[i].second;
  }
  for (j = 0; ms.size() > 0; j++) {
    o += j * *ms.begin();
    ms.erase(ms.begin());
  }
  cout << o << '\n';
}

```