#include <bits/stdc++.h>
using namespace std;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
void solve(int qq) {
  int n;
  cin >> n;
  vector<long long> a(n), cost(n);
  multiset<long long> s;
  long long sum = 0, ans = 0;
  vector<pair<long long, int> > to;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    to.push_back(make_pair(a[i], i));
  }
  for (int i = 0; i < n; i++) {
    cin >> cost[i];
  }
  sort((to).begin(), (to).end());
  int j = 0;
  long long cur = to[j].first;
  while (!s.empty() || j < n) {
    if (j < n && s.empty()) {
      cur = to[j].first;
    }
    while (j < n && to[j].first == cur) {
      s.insert(cost[to[j].second]);
      sum += cost[to[j].second];
      j++;
    }
    while (!s.empty() && ((j < n && cur < to[j].first) || j >= n)) {
      long long rem = *s.rbegin();
      sum -= *s.rbegin();
      s.erase(s.find(rem));
      ans += sum;
      cur++;
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
void solve(int qq) {
  int n;
  cin >> n;
  vector<long long> a(n), cost(n);
  multiset<long long> s;
  long long sum = 0, ans = 0;
  vector<pair<long long, int> > to;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    to.push_back(make_pair(a[i], i));
  }
  for (int i = 0; i < n; i++) {
    cin >> cost[i];
  }
  sort((to).begin(), (to).end());
  int j = 0;
  long long cur = to[j].first;
  while (!s.empty() || j < n) {
    if (j < n && s.empty()) {
      cur = to[j].first;
    }
    while (j < n && to[j].first == cur) {
      s.insert(cost[to[j].second]);
      sum += cost[to[j].second];
      j++;
    }
    while (!s.empty() && ((j < n && cur < to[j].first) || j >= n)) {
      long long rem = *s.rbegin();
      sum -= *s.rbegin();
      s.erase(s.find(rem));
      ans += sum;
      cur++;
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}

```