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;
struct node {
  long long x, y;
  bool operator<(const node &a) const { return x < a.x; }
} T[200005];
signed main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    node &a = T[i];
    cin >> a.x;
  }
  for (long long i = 0; i < n; i++) {
    node &a = T[i];
    cin >> a.y;
  }
  sort(T, T + n);
  multiset<long long> st;
  long long now = -1, ans = 0, sum = 0;
  long long mx;
  for (long long i = 0; i < n; i++) {
    while (st.size() && now < T[i].x) {
      mx = *st.rbegin();
      ans += sum - mx;
      sum -= mx;
      now++;
      st.erase(st.find(mx));
    }
    sum += T[i].y;
    st.insert(T[i].y);
    now = T[i].x;
  }
  while (st.size()) {
    mx = *st.rbegin();
    ans += sum - mx;
    sum -= mx;
    st.erase(st.find(mx));
  }
  cout << ans << endl;
}
