#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long int gcd, masks, mask, i, h, n, d, l, r, j, k, x, y, z, x1, y1, p,
      x2, y2, m, root, t, sum, size, ind;
  cin >> n;
  int a[n], b[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  vector<pair<long long int, long long int> > vp;
  for (i = 0; i < n; i++) {
    vp.push_back({a[i], b[i]});
  }
  sort(vp.begin(), vp.end());
  priority_queue<pair<long long int, long long int> > pq;
  x = vp[0].first;
  i = 0;
  long long int ans = 0;
  while (i < n || pq.size() != 0) {
    while (i < n && x == vp[i].first) {
      pq.push({vp[i].second, vp[i].first});
      i++;
    }
    ans += ((x - pq.top().second) * pq.top().first);
    pq.pop();
    if (i != n && pq.size() == 0)
      x = vp[i].first;
    else
      x++;
  }
  cout << ans;
  return 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long int gcd, masks, mask, i, h, n, d, l, r, j, k, x, y, z, x1, y1, p,
      x2, y2, m, root, t, sum, size, ind;
  cin >> n;
  int a[n], b[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  vector<pair<long long int, long long int> > vp;
  for (i = 0; i < n; i++) {
    vp.push_back({a[i], b[i]});
  }
  sort(vp.begin(), vp.end());
  priority_queue<pair<long long int, long long int> > pq;
  x = vp[0].first;
  i = 0;
  long long int ans = 0;
  while (i < n || pq.size() != 0) {
    while (i < n && x == vp[i].first) {
      pq.push({vp[i].second, vp[i].first});
      i++;
    }
    ans += ((x - pq.top().second) * pq.top().first);
    pq.pop();
    if (i != n && pq.size() == 0)
      x = vp[i].first;
    else
      x++;
  }
  cout << ans;
  return 0;
}

```