#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e5 + 1e5;
vector<long long int> v[N];
long long int path = 0;
long long int dest = 0;
vector<long long int> beta[N + 1];
double prob[N + 1];
double dis[N + 1];
vector<long long int> baap[N + 1];
long long int solve() {
  long long int n;
  cin >> n;
  long long int a[n + 1];
  long long int p[n + 1];
  unordered_map<long long int, priority_queue<long long int> > pri;
  vector<pair<long long int, long long int> > vc;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  for (long long int j = 1; j <= n; j++) {
    cin >> p[j];
    vc.push_back(make_pair(a[j], p[j]));
  }
  sort(vc.begin(), vc.end());
  priority_queue<long long int> q;
  long long int res = 0;
  long long int sum = 0;
  for (long long int i = 0; i < vc.size(); i++) {
    long long int num = vc[i].first;
    long long int val = vc[i].second;
    if (i == 0) {
      q.push(val);
      sum += val;
    } else {
      long long int diff = vc[i].first - vc[i - 1].first;
      for (long long int j = 1; j <= diff; j++) {
        if (q.empty()) break;
        sum = sum - q.top();
        res = res + sum;
        q.pop();
      }
      q.push(val);
      sum += val;
    }
  }
  while (q.size()) {
    sum = sum - q.top();
    q.pop();
    res += sum;
  }
  cout << res << endl;
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(cin.tie(NULL));
  cout.tie(NULL);
  long long int tc = 1;
  while (tc--) {
    solve();
  }
}

### 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;
const long long int N = 1e5 + 1e5;
vector<long long int> v[N];
long long int path = 0;
long long int dest = 0;
vector<long long int> beta[N + 1];
double prob[N + 1];
double dis[N + 1];
vector<long long int> baap[N + 1];
long long int solve() {
  long long int n;
  cin >> n;
  long long int a[n + 1];
  long long int p[n + 1];
  unordered_map<long long int, priority_queue<long long int> > pri;
  vector<pair<long long int, long long int> > vc;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  for (long long int j = 1; j <= n; j++) {
    cin >> p[j];
    vc.push_back(make_pair(a[j], p[j]));
  }
  sort(vc.begin(), vc.end());
  priority_queue<long long int> q;
  long long int res = 0;
  long long int sum = 0;
  for (long long int i = 0; i < vc.size(); i++) {
    long long int num = vc[i].first;
    long long int val = vc[i].second;
    if (i == 0) {
      q.push(val);
      sum += val;
    } else {
      long long int diff = vc[i].first - vc[i - 1].first;
      for (long long int j = 1; j <= diff; j++) {
        if (q.empty()) break;
        sum = sum - q.top();
        res = res + sum;
        q.pop();
      }
      q.push(val);
      sum += val;
    }
  }
  while (q.size()) {
    sum = sum - q.top();
    q.pop();
    res += sum;
  }
  cout << res << endl;
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(cin.tie(NULL));
  cout.tie(NULL);
  long long int tc = 1;
  while (tc--) {
    solve();
  }
}

```