#include <bits/stdc++.h>
#pragma GCC optimize("-Ofast")
using namespace std;
long long power(long long x, long long y, long long m) {
  long long temp;
  if (y == 0) return 1;
  temp = (power(x, y / 2, m)) % m;
  if (y % 2 == 0)
    return ((temp % m) * temp);
  else
    return ((x * temp % m) * temp % m) % m;
}
long long inv(long long x) {
  return (power(x, 1000000007 - 2, 1000000007)) % 1000000007;
}
bool comp(long long a, long long b) { return a > b; }
map<long long, long long> path;
long long finder(long long n) {
  if (n + 1 != (path.upper_bound(n)->first)) {
    return (n + 1);
  } else
    return (path.at(n) = finder(path.at(n)));
}
int32_t main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, counter = 0;
  cin >> n;
  long long arr[n], arr2[n];
  vector<pair<long long, long long>> lolz;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> arr2[i];
  }
  for (long long i = 0; i < n; i++) {
    lolz.push_back({-arr2[i], arr[i]});
  }
  sort(lolz.begin(), lolz.end());
  for (long long i = 0; i < n; i++) {
    if (path.find(lolz[i].second) == path.end()) {
      path.insert({lolz[i].second, lolz[i].second + 1});
    } else {
      long long result = finder(lolz[i].second);
      counter += (-lolz[i].second + result) * (-lolz[i].first);
      path.insert({result, result + 1});
    }
  }
  cout << counter;
  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>
#pragma GCC optimize("-Ofast")
using namespace std;
long long power(long long x, long long y, long long m) {
  long long temp;
  if (y == 0) return 1;
  temp = (power(x, y / 2, m)) % m;
  if (y % 2 == 0)
    return ((temp % m) * temp);
  else
    return ((x * temp % m) * temp % m) % m;
}
long long inv(long long x) {
  return (power(x, 1000000007 - 2, 1000000007)) % 1000000007;
}
bool comp(long long a, long long b) { return a > b; }
map<long long, long long> path;
long long finder(long long n) {
  if (n + 1 != (path.upper_bound(n)->first)) {
    return (n + 1);
  } else
    return (path.at(n) = finder(path.at(n)));
}
int32_t main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, counter = 0;
  cin >> n;
  long long arr[n], arr2[n];
  vector<pair<long long, long long>> lolz;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> arr2[i];
  }
  for (long long i = 0; i < n; i++) {
    lolz.push_back({-arr2[i], arr[i]});
  }
  sort(lolz.begin(), lolz.end());
  for (long long i = 0; i < n; i++) {
    if (path.find(lolz[i].second) == path.end()) {
      path.insert({lolz[i].second, lolz[i].second + 1});
    } else {
      long long result = finder(lolz[i].second);
      counter += (-lolz[i].second + result) * (-lolz[i].first);
      path.insert({result, result + 1});
    }
  }
  cout << counter;
  return 0;
}

```