#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using ll = long long;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
using ull = unsigned long long;
using ld = long double;
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n), t(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> t[i];
  vector<pii> b(n);
  for (int i = 0; i < n; i++) {
    b[i] = {a[i], t[i]};
  }
  sort((b).begin(), (b).end());
  ll ans = 0, res = 0, cur = -1;
  priority_queue<int> q;
  for (int i = 0; i < n;) {
    if (q.size()) {
      while (q.size() && cur < b[i].first) {
        res -= q.top();
        q.pop();
        ans += res;
        cur++;
      }
    }
    q.push(b[i].second);
    cur = b[i].first;
    res += b[i].second;
    int j = i + 1;
    while (j < n && b[j].first == cur) {
      q.push(b[j].second);
      res += b[j].second;
      j++;
    }
    res -= q.top();
    ans += res;
    q.pop();
    cur++;
    i = j;
  }
  while (q.size()) {
    res -= q.top();
    ans += res;
    q.pop();
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
using pii = pair<int, int>;
using ll = long long;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
using ull = unsigned long long;
using ld = long double;
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n), t(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> t[i];
  vector<pii> b(n);
  for (int i = 0; i < n; i++) {
    b[i] = {a[i], t[i]};
  }
  sort((b).begin(), (b).end());
  ll ans = 0, res = 0, cur = -1;
  priority_queue<int> q;
  for (int i = 0; i < n;) {
    if (q.size()) {
      while (q.size() && cur < b[i].first) {
        res -= q.top();
        q.pop();
        ans += res;
        cur++;
      }
    }
    q.push(b[i].second);
    cur = b[i].first;
    res += b[i].second;
    int j = i + 1;
    while (j < n && b[j].first == cur) {
      q.push(b[j].second);
      res += b[j].second;
      j++;
    }
    res -= q.top();
    ans += res;
    q.pop();
    cur++;
    i = j;
  }
  while (q.size()) {
    res -= q.top();
    ans += res;
    q.pop();
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

```