#include <bits/stdc++.h>
using namespace std;
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
constexpr int MAXN = 200000;
pii arr[MAXN];
int tmp[MAXN];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    cin >> arr[i].first;
    tmp[i] = arr[i].first;
  }
  for (int i = 0; i < (n); ++i) {
    cin >> arr[i].second;
  }
  sort(tmp, tmp + n);
  int cur = 0;
  set<int> isfree;
  for (int i = 0; i < (n); ++i) {
    if (tmp[i] <= cur) {
      ++cur;
    } else {
      cur = tmp[i];
    }
    isfree.insert(cur);
  }
  sort(arr, arr + n,
       [](const pii& a, const pii& b) { return a.second > b.second; });
  ll ans = 0;
  for (int i = 0; i < (n); ++i) {
    auto [v, t] = arr[i];
    auto it = isfree.lower_bound(v);
    ans += 1LL * t * (*it - v);
    isfree.erase(it);
  }
  cout << ans << "\n";
  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;
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
constexpr int MAXN = 200000;
pii arr[MAXN];
int tmp[MAXN];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    cin >> arr[i].first;
    tmp[i] = arr[i].first;
  }
  for (int i = 0; i < (n); ++i) {
    cin >> arr[i].second;
  }
  sort(tmp, tmp + n);
  int cur = 0;
  set<int> isfree;
  for (int i = 0; i < (n); ++i) {
    if (tmp[i] <= cur) {
      ++cur;
    } else {
      cur = tmp[i];
    }
    isfree.insert(cur);
  }
  sort(arr, arr + n,
       [](const pii& a, const pii& b) { return a.second > b.second; });
  ll ans = 0;
  for (int i = 0; i < (n); ++i) {
    auto [v, t] = arr[i];
    auto it = isfree.lower_bound(v);
    ans += 1LL * t * (*it - v);
    isfree.erase(it);
  }
  cout << ans << "\n";
  return 0;
}

```