#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long a[n], b[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) cin >> b[i];
  multiset<long long> ms;
  map<long long, vector<long long>> mp;
  for (long long i = 0; i < n; i++) mp[a[i]].push_back(b[i]);
  auto it = mp.begin();
  long long ans = 0, sum = 0;
  while (it != mp.end()) {
    long long pos = it->first;
    auto nxt = it;
    nxt++;
    long long nxt_pos = nxt == mp.end() ? (long long)(1e17 + 10) : nxt->first;
    for (auto i : (it->second)) {
      ms.insert(-i);
      sum += i;
    }
    sum += *ms.begin();
    ms.erase(ms.begin());
    long long cur_pos = pos + 1;
    while (!ms.empty()) {
      if (cur_pos == nxt_pos) {
        ans += sum * (nxt_pos - pos);
        break;
      } else {
        auto itr = ms.begin();
        long long ind = *itr;
        ans -= ind * (cur_pos - pos);
        sum += ind;
        ms.erase(itr);
        cur_pos++;
      }
    }
    it++;
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long a[n], b[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) cin >> b[i];
  multiset<long long> ms;
  map<long long, vector<long long>> mp;
  for (long long i = 0; i < n; i++) mp[a[i]].push_back(b[i]);
  auto it = mp.begin();
  long long ans = 0, sum = 0;
  while (it != mp.end()) {
    long long pos = it->first;
    auto nxt = it;
    nxt++;
    long long nxt_pos = nxt == mp.end() ? (long long)(1e17 + 10) : nxt->first;
    for (auto i : (it->second)) {
      ms.insert(-i);
      sum += i;
    }
    sum += *ms.begin();
    ms.erase(ms.begin());
    long long cur_pos = pos + 1;
    while (!ms.empty()) {
      if (cur_pos == nxt_pos) {
        ans += sum * (nxt_pos - pos);
        break;
      } else {
        auto itr = ms.begin();
        long long ind = *itr;
        ans -= ind * (cur_pos - pos);
        sum += ind;
        ms.erase(itr);
        cur_pos++;
      }
    }
    it++;
  }
  cout << ans;
  return 0;
}

```