#include <bits/stdc++.h>
using namespace ::std;
vector<pair<unsigned long long, unsigned long long> > v;
unsigned long long a[1000000];
int main() {
  unsigned long long i, j, k, n, l;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) {
    cin >> l;
    v.push_back(make_pair(a[i], l));
  }
  sort(v.begin(), v.end());
  multiset<unsigned long long, greater<unsigned long long> > s;
  unsigned long long ans = 0, prevsum = 0;
  for (i = 0; i < n;) {
    j = i + 1;
    s.insert(v[i].second);
    unsigned long long sum = v[i].second + prevsum;
    while (j < n && v[j].first == v[i].first) {
      s.insert(v[j].second);
      sum += v[j].second;
      j++;
    }
    unsigned long long gap = 10000000000000;
    unsigned long long num = s.size();
    if (j != n) {
      gap = v[j].first - v[i].first;
    }
    sum -= (*s.begin());
    s.erase(s.begin());
    k = 0;
    while (k < min(num - 1, gap - 1)) {
      ans = ans + (k + 1) * (*s.begin());
      sum -= (*s.begin());
      s.erase(s.begin());
      k++;
    }
    if (j != n) ans += (sum * (gap));
    prevsum = sum;
    i = j;
  }
  k = 0;
  while (!s.empty()) {
    ans = ans + (k + 1) * (*s.begin());
    s.erase(s.begin());
    k++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<pair<unsigned long long, unsigned long long> > v;
unsigned long long a[1000000];
int main() {
  unsigned long long i, j, k, n, l;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) {
    cin >> l;
    v.push_back(make_pair(a[i], l));
  }
  sort(v.begin(), v.end());
  multiset<unsigned long long, greater<unsigned long long> > s;
  unsigned long long ans = 0, prevsum = 0;
  for (i = 0; i < n;) {
    j = i + 1;
    s.insert(v[i].second);
    unsigned long long sum = v[i].second + prevsum;
    while (j < n && v[j].first == v[i].first) {
      s.insert(v[j].second);
      sum += v[j].second;
      j++;
    }
    unsigned long long gap = 10000000000000;
    unsigned long long num = s.size();
    if (j != n) {
      gap = v[j].first - v[i].first;
    }
    sum -= (*s.begin());
    s.erase(s.begin());
    k = 0;
    while (k < min(num - 1, gap - 1)) {
      ans = ans + (k + 1) * (*s.begin());
      sum -= (*s.begin());
      s.erase(s.begin());
      k++;
    }
    if (j != n) ans += (sum * (gap));
    prevsum = sum;
    i = j;
  }
  k = 0;
  while (!s.empty()) {
    ans = ans + (k + 1) * (*s.begin());
    s.erase(s.begin());
    k++;
  }
  cout << ans << endl;
  return 0;
}

```