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 N = 2e5 + 5;
int n;
pair<long long, long long> a[N];
bool cmp(const pair<long long, long long>& a,
         const pair<long long, long long>& b) {
  return (a.first < b.first) || (a.first == b.first && a.second > b.second);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i].first;
  for (int i = 1; i <= n; ++i) cin >> a[i].second;
  sort(a + 1, a + 1 + n, cmp);
  vector<pair<int, vector<int> > > g;
  for (int i = 1; i <= n;) {
    int j = i;
    pair<int, vector<int> > tmp;
    tmp.first = a[i].first;
    while (j <= n && a[j].first == a[i].first) {
      tmp.second.push_back(a[j].second);
      ++j;
    }
    g.push_back(tmp);
    i = j;
  }
  int m = int(g.size()), j = 0;
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >,
                 greater<pair<long long, long long> > >
      heap;
  long long ans = 0;
  for (int cur = 1; j < m || !heap.empty(); ++cur) {
    while (j < m && g[j].first <= cur) {
      for (int k = 0; k < int(g[j].second.size()); ++k) {
        heap.push(pair<long long, long long>(-g[j].second[k], g[j].first));
      }
      ++j;
    }
    if (heap.empty()) {
      cur = g[j].first - 1;
      continue;
    }
    pair<long long, long long> u = heap.top();
    heap.pop();
    long long res = -(1LL * cur - u.second) * u.first;
    ans += res;
  }
  cout << ans;
  return 0;
}
