#include <bits/stdc++.h>
using namespace std;
long long sol = 0;
int pMax = 0;
void findNextAvailable(set<int>& s, map<int, int>& m, int pos) {
  if (m.find(pos + 1) == m.end()) s.insert(pos + 1);
}
void insertMap(set<int>& s, map<int, int>& m, pair<int, int> val) {
  if (m.find(val.first) == m.end()) {
    m[val.first]++;
    if (s.find(val.first) != s.end()) {
      s.erase(s.find(val.first));
    }
    findNextAvailable(s, m, val.first);
  } else {
    auto it = s.upper_bound(val.first);
    int pos = *it;
    sol += (0ll + pos - val.first) * val.second;
    s.erase(it);
    findNextAvailable(s, m, pos);
    m[pos]++;
  }
}
int cmp(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second > b.second;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end(), cmp);
  set<int> s;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    insertMap(s, m, a[i]);
  }
  cout << sol << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 sol = 0;
int pMax = 0;
void findNextAvailable(set<int>& s, map<int, int>& m, int pos) {
  if (m.find(pos + 1) == m.end()) s.insert(pos + 1);
}
void insertMap(set<int>& s, map<int, int>& m, pair<int, int> val) {
  if (m.find(val.first) == m.end()) {
    m[val.first]++;
    if (s.find(val.first) != s.end()) {
      s.erase(s.find(val.first));
    }
    findNextAvailable(s, m, val.first);
  } else {
    auto it = s.upper_bound(val.first);
    int pos = *it;
    sol += (0ll + pos - val.first) * val.second;
    s.erase(it);
    findNextAvailable(s, m, pos);
    m[pos]++;
  }
}
int cmp(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second > b.second;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end(), cmp);
  set<int> s;
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    insertMap(s, m, a[i]);
  }
  cout << sol << endl;
  return 0;
}

```