#include <bits/stdc++.h>
using namespace std;
long long int godram;
int main() {
  long long int n;
  cin >> n;
  long long int b[n + 1];
  long long int g[n + 1];
  long long int a[n + 1];
  long long int i = 1;
  while (i <= n) {
    cin >> a[i];
    long long int p = a[i];
    g[i] = p;
    i++;
  }
  sort(a + 1, a + n + 1);
  multiset<long long int> w;
  long long int max = 0;
  i = 1;
  while (i <= n) {
    if (a[i] <= max) {
      b[i] = max + 1;
      max++;
    } else {
      b[i] = a[i];
      max = a[i];
    }
    w.insert(b[i]);
    i++;
  }
  vector<pair<long long int, long long int>> c;
  i = 1;
  while (i <= n) {
    long long int x;
    cin >> x;
    pair<long long int, long long int> ram = {x, i};
    c.push_back(ram);
    i++;
  }
  sort(c.begin(), c.end());
  i = n - 1;
  while (i >= 0) {
    long long int cost = c[i].first;
    long long int index = c[i].second;
    long long int actual = g[index];
    auto it = w.upper_bound(actual - 1);
    long long int real = *it;
    w.erase(w.find(real));
    godram += abs(actual - real) * cost;
    i--;
  }
  cout << godram;
  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;
long long int godram;
int main() {
  long long int n;
  cin >> n;
  long long int b[n + 1];
  long long int g[n + 1];
  long long int a[n + 1];
  long long int i = 1;
  while (i <= n) {
    cin >> a[i];
    long long int p = a[i];
    g[i] = p;
    i++;
  }
  sort(a + 1, a + n + 1);
  multiset<long long int> w;
  long long int max = 0;
  i = 1;
  while (i <= n) {
    if (a[i] <= max) {
      b[i] = max + 1;
      max++;
    } else {
      b[i] = a[i];
      max = a[i];
    }
    w.insert(b[i]);
    i++;
  }
  vector<pair<long long int, long long int>> c;
  i = 1;
  while (i <= n) {
    long long int x;
    cin >> x;
    pair<long long int, long long int> ram = {x, i};
    c.push_back(ram);
    i++;
  }
  sort(c.begin(), c.end());
  i = n - 1;
  while (i >= 0) {
    long long int cost = c[i].first;
    long long int index = c[i].second;
    long long int actual = g[index];
    auto it = w.upper_bound(actual - 1);
    long long int real = *it;
    w.erase(w.find(real));
    godram += abs(actual - real) * cost;
    i--;
  }
  cout << godram;
  return 0;
}

```