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;
long long res = 0;
vector<pair<int, int> > v;
map<int, int> final;
int T, n, a[1 << 18], t[1 << 18];
void insert(int x) {
  auto u = final.lower_bound(x + 1), v = u;
  v--;
  if (x == v->second + 1 && x + 1 == u->first)
    final[v->first] = u->second, final.erase(u);
  else if (x == v->second + 1)
    final[v->first] = x;
  else if (x + 1 == u->first)
    final[x] = u->second, final.erase(u);
  else
    final[x] = x;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < n; i++) scanf("%d", t + i);
  for (int i = 0; i < n; i++) v.push_back(pair<int, int>(t[i], a[i]));
  sort(v.begin(), v.end(), greater<pair<int, int> >()), final[0] = 0;
  for (int i = 0; i < v.size(); i++) {
    int cur;
    auto u = final.lower_bound(v[i].second + 1), k = u;
    k--, cur = max(v[i].second, k->second + 1);
    res += 1LL * (cur - v[i].second) * v[i].first, insert(cur);
  }
  cout << res << endl;
}
