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 = 200000;
int main() {
  struct ele {
    int c;
    int t;
  } a[N];
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i].c;
  for (int i = 0; i < n; ++i) cin >> a[i].t;
  sort(a, a + n, [&](ele a, ele b) { return a.c < b.c; });
  pair<int, int> baze[N];
  baze[0].first = 0;
  int bs = 0;
  for (int i = 1; i < n; ++i)
    if (a[i].c != a[i - 1].c) {
      baze[bs++].second = i;
      baze[bs].first = i;
    }
  baze[bs++].second = n;
  int p = 0;
  int c;
  multiset<int, greater<int> > S;
  long long ans = 0;
  long long sum = 0;
  for (p = 0; p < bs; p++) {
    c = a[baze[p].first].c;
    for (int i = baze[p].first; i < baze[p].second; ++i) {
      S.insert(a[i].t);
      sum += a[i].t;
    }
    while (!S.empty()) {
      sum -= *S.begin();
      S.erase(S.begin());
      ans += sum;
      c++;
      if (p + 1 < bs && c == a[baze[p + 1].first].c) break;
    }
  }
  cout << ans;
}
