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;
const int maxn = 2 * (1e5) + 5;
int a[maxn];
long long int t[maxn];
map<int, vector<long long int> > oc;
vector<pair<int, vector<long long int> > > ocv;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  vector<long long int> fst;
  for (int i = 0; i < n; i++) {
    if (oc.find(a[i]) == oc.end()) {
      fst.push_back(t[i]);
      oc.insert({a[i], fst});
      fst.clear();
    } else {
      oc[a[i]].push_back(t[i]);
    }
  }
  for (map<int, vector<long long int> >::iterator iter = oc.begin();
       iter != oc.end(); iter++) {
    ocv.push_back((*iter));
  }
  int ocvsize = ocv.size();
  priority_queue<long long int> pq;
  long long int ans = 0;
  long long int sumcur = 0;
  for (int i = 0; i < ocvsize; i++) {
    int av;
    for (vector<long long int>::iterator iter = ocv[i].second.begin();
         iter != ocv[i].second.end(); iter++) {
      pq.push((*iter));
      sumcur += (*iter);
    }
    if (i != ocvsize - 1) {
      av = min((int)pq.size(), ocv[i + 1].first - ocv[i].first);
    } else {
      av = pq.size();
    }
    for (int i = 0; i < av; i++) {
      ans = ans + (long long)i * (long long)pq.top();
      sumcur -= (long long)pq.top();
      pq.pop();
    }
    if (i != ocvsize - 1) {
      ans += (sumcur) * (long long)(ocv[i + 1].first - ocv[i].first);
    }
  }
  cout << ans << endl;
  return 0;
}
