#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
const long long N = 100011;
void solve() {
  long long n, ans = 0;
  cin >> n;
  pair<long long, long long> a[n + 17];
  multiset<long long> s;
  for (long long i = 1; i <= n; i++) cin >> a[i].first;
  for (long long i = 1; i <= n; i++) cin >> a[i].second;
  sort(a + 1, a + n + 1);
  long long sum = 0;
  long long mn = 0;
  a[n + 1].first = -1;
  for (long long ind = 1; ind <= n; ind++) {
    mn += 1;
    if (s.empty()) mn = a[ind].first;
    long long j = ind;
    while (a[j].first == mn) {
      s.insert(a[j].second);
      sum += a[j].second;
      j += 1;
    }
    long long mx = *(--s.end());
    sum -= mx;
    s.erase(s.find(mx));
    ans += sum;
    ind = j - 1;
  }
  while (s.size()) {
    int mx = *(--s.end());
    sum -= mx;
    s.erase(s.find(mx));
    ans += sum;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
    cout << endl;
  }
}

### 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 mod = 1e9 + 7;
const long long N = 100011;
void solve() {
  long long n, ans = 0;
  cin >> n;
  pair<long long, long long> a[n + 17];
  multiset<long long> s;
  for (long long i = 1; i <= n; i++) cin >> a[i].first;
  for (long long i = 1; i <= n; i++) cin >> a[i].second;
  sort(a + 1, a + n + 1);
  long long sum = 0;
  long long mn = 0;
  a[n + 1].first = -1;
  for (long long ind = 1; ind <= n; ind++) {
    mn += 1;
    if (s.empty()) mn = a[ind].first;
    long long j = ind;
    while (a[j].first == mn) {
      s.insert(a[j].second);
      sum += a[j].second;
      j += 1;
    }
    long long mx = *(--s.end());
    sum -= mx;
    s.erase(s.find(mx));
    ans += sum;
    ind = j - 1;
  }
  while (s.size()) {
    int mx = *(--s.end());
    sum -= mx;
    s.erase(s.find(mx));
    ans += sum;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
    cout << endl;
  }
}

```