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.
#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) {
  if (a < b) return b;
  return a;
}
void solve() {
  long long a, b, n, p;
  cin >> n;
  unordered_map<long long, int> umap;
  vector<long long> v, t;
  vector<pair<long long, long long> > vec;
  set<long long> s;
  for (int i = 0; i < n; i++) {
    long long te;
    cin >> te;
    v.push_back(te);
  }
  for (int i = 0; i < n; i++) {
    long long te;
    cin >> te;
    t.push_back(te);
  }
  for (int i = 0; i < n; i++) {
    vec.push_back(make_pair(t[i], v[i]));
  }
  sort(v.begin(), v.end());
  long long g = 0;
  for (int i = 0; i < n; i++) {
    if (v[i] > g) {
      g = v[i];
      s.insert(g);
    } else {
      g++;
      s.insert(g);
    }
  }
  sort(vec.begin(), vec.end());
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    auto e = s.lower_bound(vec[i].second);
    ans += vec[i].first * (*e - vec[i].second);
    s.erase(e);
  }
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}
