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;
vector<pair<long long, long long> > v;
long long A[1000006], cost[1000006], res;
map<long long, long long> par;
long long rep(long long x) {
  if (!par[x]) return par[x] = x;
  if (par[x] == x) return x;
  return par[x] = rep(par[x]);
}
void join(long long x, long long y) {
  long long a = rep(x);
  long long b = rep(y);
  if (a != b) {
    par[b] = a;
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    par[A[i]] = A[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> cost[i];
  }
  for (int i = 1; i <= n; i++) {
    v.push_back({-cost[i], A[i]});
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    long long x = v[i].second;
    res += (rep(x) - x) * -v[i].first;
    join(rep(rep(x) + 1), rep(x));
  }
  cout << res << endl;
  return 0;
}
