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 = 2e5 + 10;
struct Node {
  long long num, t;
  bool operator<(const Node &b) const {
    if (t == b.t) return num < b.num;
    return t > b.t;
  }
} a[maxn];
map<int, int> pre;
int findx(int x) { return pre[x] == 0 ? x : pre[x] = findx(pre[x]); }
void merge(int x, int y) {
  int i = findx(x), j = findx(y);
  if (i != j) pre[i] = j;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].num;
  for (int i = 1; i <= n; i++) cin >> a[i].t;
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int res = findx(a[i].num);
    if (res == a[i].num)
      merge(res, res + 1);
    else {
      merge(res, res + 1);
      ans += 1ll * (res - a[i].num) * a[i].t;
    }
  }
  cout << ans << "\n";
  return 0;
}
