#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const ll inf = 1e18 + 7;
const ll mod = 1000000007;
const int mxn = 1e9 + 7;
const ld eps = 1e-9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n), t(n);
  vector<pair<int, int>> v;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> t[i];
    v.push_back({a[i], t[i]});
  }
  sort(v.begin(), v.end());
  ll ans = 0, c = v[0].second, pref = v[0].first;
  multiset<int> ms{v[0].second};
  for (int i = 1; i < n; ++i) {
    while (v[i].first > pref && !ms.empty()) {
      ++pref;
      c -= *(--ms.end());
      ms.erase(--ms.end());
      ans += c;
    }
    ms.insert(v[i].second);
    c += v[i].second;
    pref = v[i].first;
  }
  while (!ms.empty()) {
    c -= *(--ms.end());
    ms.erase(--ms.end());
    ans += c;
  }
  cout << ans << '\n';
}

### 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;
using ll = long long;
using ld = long double;
const ll inf = 1e18 + 7;
const ll mod = 1000000007;
const int mxn = 1e9 + 7;
const ld eps = 1e-9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n), t(n);
  vector<pair<int, int>> v;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> t[i];
    v.push_back({a[i], t[i]});
  }
  sort(v.begin(), v.end());
  ll ans = 0, c = v[0].second, pref = v[0].first;
  multiset<int> ms{v[0].second};
  for (int i = 1; i < n; ++i) {
    while (v[i].first > pref && !ms.empty()) {
      ++pref;
      c -= *(--ms.end());
      ms.erase(--ms.end());
      ans += c;
    }
    ms.insert(v[i].second);
    c += v[i].second;
    pref = v[i].first;
  }
  while (!ms.empty()) {
    c -= *(--ms.end());
    ms.erase(--ms.end());
    ans += c;
  }
  cout << ans << '\n';
}

```