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 x, p;
} a[maxn << 1];
bool cmp(node a, node b) {
  if (a.p == b.p)
    return a.x < b.x;
  else
    return a.p > b.p;
}
map<long long, long long> f;
long long Find(long long x) { return f[x] == 0 ? x : f[x] = Find(f[x]); }
void Union(long long x, long long y) {
  x = Find(x), y = Find(y);
  if (x != y) f[x] = y;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].x;
  for (int i = 1; i <= n; i++) cin >> a[i].p;
  sort(a + 1, a + n + 1, cmp);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long tmp = Find(a[i].x);
    ans += (tmp - a[i].x) * a[i].p;
    Union(tmp, tmp + 1);
  }
  cout << ans << endl;
}
