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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << endl;
  err(++it, args...);
}
const int mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  vector<pair<int, int> > vect(n);
  for (int i = 0; i < n; i++) cin >> vect[i].first;
  for (int i = 0; i < n; i++) cin >> vect[i].second;
  sort(vect.begin(), vect.end());
  multiset<int> st;
  long long i = 0, temp, sum = 0, ans = 0;
  while (true) {
    if (st.size() == 0 && i == n) return cout << ans, 0;
    if (st.size() == 0) temp = vect[i].first;
    for (; vect[i].first == temp; i++) {
      st.insert(vect[i].second);
      sum += vect[i].second;
    }
    sum -= (*st.rbegin());
    ans += sum;
    st.erase(--st.end());
    temp++;
  }
}
