#include <bits/stdc++.h>
using namespace std;
inline long long add(long long a, long long b) {
  return ((a % 1000000007LL) + (b % 1000000007LL)) % 1000000007LL;
}
inline long long mul(long long a, long long b) {
  return ((a % 1000000007LL) * (b % 1000000007LL)) % 1000000007LL;
}
inline long long sub(long long a, long long b) {
  return ((a % 1000000007LL) - (b % 1000000007LL) + 1000000007LL) %
         1000000007LL;
}
void Flash() {
  long long n;
  cin >> n;
  pair<long long, long long> a[n];
  map<long long, vector<long long>> mpp;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first;
  }
  for (long long i = 0; i < n; i++) {
    cin >> a[i].second;
    mpp[a[i].first].push_back(a[i].second);
  }
  priority_queue<long long> pq;
  long long tSum = 0;
  long long ans = 0;
  vector<long long> temp;
  for (auto it : mpp) {
    for (auto itt : it.second) {
      pq.push(itt);
      tSum += itt;
    }
    tSum -= pq.top();
    pq.pop();
    ans += tSum;
    if (!pq.empty()) {
      mpp.insert(make_pair(it.first + 1, temp));
    }
  }
  cout << ans << "\n";
}
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  ;
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    Flash();
  }
  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;
inline long long add(long long a, long long b) {
  return ((a % 1000000007LL) + (b % 1000000007LL)) % 1000000007LL;
}
inline long long mul(long long a, long long b) {
  return ((a % 1000000007LL) * (b % 1000000007LL)) % 1000000007LL;
}
inline long long sub(long long a, long long b) {
  return ((a % 1000000007LL) - (b % 1000000007LL) + 1000000007LL) %
         1000000007LL;
}
void Flash() {
  long long n;
  cin >> n;
  pair<long long, long long> a[n];
  map<long long, vector<long long>> mpp;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first;
  }
  for (long long i = 0; i < n; i++) {
    cin >> a[i].second;
    mpp[a[i].first].push_back(a[i].second);
  }
  priority_queue<long long> pq;
  long long tSum = 0;
  long long ans = 0;
  vector<long long> temp;
  for (auto it : mpp) {
    for (auto itt : it.second) {
      pq.push(itt);
      tSum += itt;
    }
    tSum -= pq.top();
    pq.pop();
    ans += tSum;
    if (!pq.empty()) {
      mpp.insert(make_pair(it.first + 1, temp));
    }
  }
  cout << ans << "\n";
}
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  ;
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    Flash();
  }
  return 0;
}

```