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;
int n, t;
struct node {
  int data;
  long long cost;
  bool operator<(const node &b) const {
    if (cost == b.cost)
      return data < b.data;
    else
      return cost > b.cost;
  }
} a[3000000];
map<int, int> fa;
int find(int a) {
  if (fa[a] == 0)
    return a;
  else
    return fa[a] = find(fa[a]);
}
void unite(int x, int y) {
  x = find(x);
  y = find(y);
  if (x != y) fa[x] = y;
}
int main() {
  cin >> n;
  fa.clear();
  for (int i = 0; i < n; ++i) scanf("%d", &a[i].data);
  for (int i = 0; i < n; ++i) scanf("%lld", &a[i].cost);
  sort(a, a + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int x = find(a[i].data);
    ans += a[i].cost * (x - a[i].data);
    unite(x, x + 1);
  }
  cout << ans << endl;
}
