#include <bits/stdc++.h>
using namespace std;
int n;
int a[200005];
int t[200005];
vector<pair<int, int> > v;
priority_queue<int> pq;
bool comp2(pair<int, int>& a, pair<int, int>& b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first < b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  for (int i = 0; i < n; i++) {
    v.push_back({a[i], t[i]});
  }
  sort(v.begin(), v.end(), comp2);
  int cur = 0;
  int inque = 0;
  long long sum = 0;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (pq.empty()) {
      if (v[i].first == v[i + 1].first) {
        inque = v[i].first + 1;
        while (v[i].first == v[i + 1].first) {
          pq.push(v[i + 1].second);
          sum += v[i + 1].second;
          i++;
        }
        ans += sum;
      } else {
        continue;
      }
    } else {
      if (inque == v[i].first) {
        pq.push(v[i].second);
        sum += v[i].second;
        continue;
      }
      while (!pq.empty() && inque < v[i].first) {
        int top = pq.top();
        pq.pop();
        sum -= top;
        ans += sum;
        inque++;
      }
      if (!pq.empty()) {
        sum += v[i].second;
        pq.push(v[i].second);
      } else {
        if (v[i].first == v[i + 1].first) {
          inque = v[i].first + 1;
          while (v[i].first == v[i + 1].first) {
            pq.push(v[i + 1].second);
            sum += v[i + 1].second;
            i++;
          }
          ans += sum;
        }
      }
    }
  }
  while (!pq.empty()) {
    sum -= pq.top();
    pq.pop();
    ans += sum;
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int n;
int a[200005];
int t[200005];
vector<pair<int, int> > v;
priority_queue<int> pq;
bool comp2(pair<int, int>& a, pair<int, int>& b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first < b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  for (int i = 0; i < n; i++) {
    v.push_back({a[i], t[i]});
  }
  sort(v.begin(), v.end(), comp2);
  int cur = 0;
  int inque = 0;
  long long sum = 0;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (pq.empty()) {
      if (v[i].first == v[i + 1].first) {
        inque = v[i].first + 1;
        while (v[i].first == v[i + 1].first) {
          pq.push(v[i + 1].second);
          sum += v[i + 1].second;
          i++;
        }
        ans += sum;
      } else {
        continue;
      }
    } else {
      if (inque == v[i].first) {
        pq.push(v[i].second);
        sum += v[i].second;
        continue;
      }
      while (!pq.empty() && inque < v[i].first) {
        int top = pq.top();
        pq.pop();
        sum -= top;
        ans += sum;
        inque++;
      }
      if (!pq.empty()) {
        sum += v[i].second;
        pq.push(v[i].second);
      } else {
        if (v[i].first == v[i + 1].first) {
          inque = v[i].first + 1;
          while (v[i].first == v[i + 1].first) {
            pq.push(v[i + 1].second);
            sum += v[i + 1].second;
            i++;
          }
          ans += sum;
        }
      }
    }
  }
  while (!pq.empty()) {
    sum -= pq.top();
    pq.pop();
    ans += sum;
  }
  cout << ans;
  return 0;
}

```