#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 23;
int n, m, T;
int vis[203];
pair<int, int> a[maxn];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].first);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].second);
  sort(a + 1, a + n + 1);
  long long ans = 00;
  multiset<int> s;
  long long tot = 0;
  for (int i = 1; i <= n; i++) {
    if (s.size() == 0 && a[i].first != a[i - 1].first) {
      continue;
    } else if (s.size() == 0 && a[i].first == a[i - 1].first) {
      s.insert(a[i].second);
      s.insert(a[i - 1].second);
      tot = a[i].second + a[i - 1].second;
    } else if (s.size() >= 1) {
      int d = a[i].first - a[i - 1].first;
      while (d > 0 && s.size() > 0) {
        auto it = --s.end();
        tot -= *it;
        ans += tot;
        d--;
        s.erase(it);
      }
      if (s.size()) {
        s.insert(a[i].second);
        tot += a[i].second;
      }
    }
  }
  while (s.size()) {
    auto it = --s.end();
    tot -= *it;
    s.erase(it);
    ans += tot;
  }
  printf("%lld", ans);
}

### 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;
const int maxn = 2e5 + 23;
int n, m, T;
int vis[203];
pair<int, int> a[maxn];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].first);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].second);
  sort(a + 1, a + n + 1);
  long long ans = 00;
  multiset<int> s;
  long long tot = 0;
  for (int i = 1; i <= n; i++) {
    if (s.size() == 0 && a[i].first != a[i - 1].first) {
      continue;
    } else if (s.size() == 0 && a[i].first == a[i - 1].first) {
      s.insert(a[i].second);
      s.insert(a[i - 1].second);
      tot = a[i].second + a[i - 1].second;
    } else if (s.size() >= 1) {
      int d = a[i].first - a[i - 1].first;
      while (d > 0 && s.size() > 0) {
        auto it = --s.end();
        tot -= *it;
        ans += tot;
        d--;
        s.erase(it);
      }
      if (s.size()) {
        s.insert(a[i].second);
        tot += a[i].second;
      }
    }
  }
  while (s.size()) {
    auto it = --s.end();
    tot -= *it;
    s.erase(it);
    ans += tot;
  }
  printf("%lld", ans);
}

```