#include <bits/stdc++.h>
using namespace std;
long long lim = 3e16 + 5;
const int mod = 1e9 + 7;
long long n, m, t, k;
long long Power(long long a, long long b) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long ans = Power((a * a) % mod, b / 2);
  if (b & 1) return (a * ans) % mod;
  return ans % mod;
}
long long Inv(long long a) { return Power(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<long long, long long> > v(n);
  for (int i = 0; i < n; i++) cin >> v[i].first;
  for (int i = 0; i < n; i++) cin >> v[i].second;
  sort(v.begin(), v.end());
  int i = 0;
  long long ans = 0;
  while (i < n) {
    int j = i + 1;
    while (j < n && v[j].first - v[i].first < j - i) j++;
    j--;
    vector<vector<long long> > x(j - i + 1);
    long long k = i;
    while (k <= j) {
      x[v[k].first - v[i].first].push_back(v[k].second);
      k++;
    }
    set<int> Set;
    map<int, int> mp;
    k = i;
    long long tmp = 0;
    for (int k = 0; k < x.size(); k++) {
      for (int p = 0; p < x[k].size(); p++) {
        Set.insert(x[k][p]);
        mp[x[k][p]]++;
        tmp += x[k][p];
      }
      auto it = Set.rbegin();
      tmp -= *it;
      mp[*it]--;
      if (mp[*it] == 0) Set.erase(*it);
      ans += tmp;
    }
    i = j + 1;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long lim = 3e16 + 5;
const int mod = 1e9 + 7;
long long n, m, t, k;
long long Power(long long a, long long b) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long ans = Power((a * a) % mod, b / 2);
  if (b & 1) return (a * ans) % mod;
  return ans % mod;
}
long long Inv(long long a) { return Power(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<long long, long long> > v(n);
  for (int i = 0; i < n; i++) cin >> v[i].first;
  for (int i = 0; i < n; i++) cin >> v[i].second;
  sort(v.begin(), v.end());
  int i = 0;
  long long ans = 0;
  while (i < n) {
    int j = i + 1;
    while (j < n && v[j].first - v[i].first < j - i) j++;
    j--;
    vector<vector<long long> > x(j - i + 1);
    long long k = i;
    while (k <= j) {
      x[v[k].first - v[i].first].push_back(v[k].second);
      k++;
    }
    set<int> Set;
    map<int, int> mp;
    k = i;
    long long tmp = 0;
    for (int k = 0; k < x.size(); k++) {
      for (int p = 0; p < x[k].size(); p++) {
        Set.insert(x[k][p]);
        mp[x[k][p]]++;
        tmp += x[k][p];
      }
      auto it = Set.rbegin();
      tmp -= *it;
      mp[*it]--;
      if (mp[*it] == 0) Set.erase(*it);
      ans += tmp;
    }
    i = j + 1;
  }
  cout << ans << endl;
  return 0;
}

```