#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
int ax[N];
int t[N];
unordered_map<int, vector<int>> times;
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ax[i];
  }
  vector<int> a;
  for (int i = 0, t; i < n; i++) {
    cin >> t;
    if (times[ax[i]].size() == 0) a.push_back(ax[i]);
    times[ax[i]].push_back(t);
  }
  priority_queue<int> pq;
  sort(a.begin(), a.end());
  long long ans = 0;
  long long qsum = 0;
  for (int i = 0; i < a.size(); i++) {
    ans += qsum;
    for (int j : times[a[i]]) {
      pq.push(j);
      qsum += j;
    }
    qsum -= pq.top();
    pq.pop();
    long long start = a[i] + 1;
    while (!pq.empty() && (i == a.size() - 1 || a[i + 1] > start)) {
      ans += qsum;
      qsum -= pq.top();
      pq.pop();
      start++;
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  getchar();
  getchar();
}

### Prompt

Generate a cpp solution 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;
const int N = 200000;
int ax[N];
int t[N];
unordered_map<int, vector<int>> times;
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ax[i];
  }
  vector<int> a;
  for (int i = 0, t; i < n; i++) {
    cin >> t;
    if (times[ax[i]].size() == 0) a.push_back(ax[i]);
    times[ax[i]].push_back(t);
  }
  priority_queue<int> pq;
  sort(a.begin(), a.end());
  long long ans = 0;
  long long qsum = 0;
  for (int i = 0; i < a.size(); i++) {
    ans += qsum;
    for (int j : times[a[i]]) {
      pq.push(j);
      qsum += j;
    }
    qsum -= pq.top();
    pq.pop();
    long long start = a[i] + 1;
    while (!pq.empty() && (i == a.size() - 1 || a[i + 1] > start)) {
      ans += qsum;
      qsum -= pq.top();
      pq.pop();
      start++;
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  getchar();
  getchar();
}

```