#include <bits/stdc++.h>
using namespace std;
long long int expo(long long int n) {
  long long int t;
  if (n == 0)
    return 1;
  else if (n == 1)
    return 2;
  t = expo(n / 2);
  if (n % 2) {
    return t * t * 2;
  } else
    return t * t;
}
long long int exponentMod(long long int A, long long int B, long long int C) {
  if (A == 0) return 0;
  if (B == 0) return 1;
  long long int y;
  if (B % 2 == 0) {
    y = exponentMod(A, B / 2, C);
    y = (y * y) % C;
  } else {
    y = A % C;
    y = (y * exponentMod(A, B - 1, C) % C) % C;
  }
  return (long long int)((y + C) % C);
}
int main() {
  long long int n;
  cin >> n;
  pair<long long int, long long int> p[n + 1];
  for (int i = 0; i < n; i++) cin >> p[i].first;
  for (int i = 0; i < n; i++) cin >> p[i].second;
  sort(p, p + n);
  priority_queue<long long int> m;
  long long int i = 0;
  long long int cur = p[i].first;
  long long int sum = 0, ans = 0;
  while (i < n || m.size()) {
    if (m.size() == 0) cur = p[i].first;
    while (i < n && cur == p[i].first) {
      m.push(p[i].second);
      sum += p[i].second;
      i++;
    }
    sum -= m.top();
    m.pop();
    ans += sum;
    cur++;
  }
  cout << ans << "\n";
}

### Prompt

Develop a solution in cpp to the problem described below:

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 expo(long long int n) {
  long long int t;
  if (n == 0)
    return 1;
  else if (n == 1)
    return 2;
  t = expo(n / 2);
  if (n % 2) {
    return t * t * 2;
  } else
    return t * t;
}
long long int exponentMod(long long int A, long long int B, long long int C) {
  if (A == 0) return 0;
  if (B == 0) return 1;
  long long int y;
  if (B % 2 == 0) {
    y = exponentMod(A, B / 2, C);
    y = (y * y) % C;
  } else {
    y = A % C;
    y = (y * exponentMod(A, B - 1, C) % C) % C;
  }
  return (long long int)((y + C) % C);
}
int main() {
  long long int n;
  cin >> n;
  pair<long long int, long long int> p[n + 1];
  for (int i = 0; i < n; i++) cin >> p[i].first;
  for (int i = 0; i < n; i++) cin >> p[i].second;
  sort(p, p + n);
  priority_queue<long long int> m;
  long long int i = 0;
  long long int cur = p[i].first;
  long long int sum = 0, ans = 0;
  while (i < n || m.size()) {
    if (m.size() == 0) cur = p[i].first;
    while (i < n && cur == p[i].first) {
      m.push(p[i].second);
      sum += p[i].second;
      i++;
    }
    sum -= m.top();
    m.pop();
    ans += sum;
    cur++;
  }
  cout << ans << "\n";
}

```