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 + 5;
struct cate {
  long long num, t;
  bool operator<(const cate& other) const { return t > other.t; }
} a[maxn];
int n;
set<long long> has;
map<long long, long long> fa;
long long find(long long x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].num;
    fa[a[i].num] = 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) {
    long long cur = a[i].num;
    if (has.find(cur) != has.end()) {
      cur = find(cur);
      cur += 1;
      fa[cur] = cur;
      ans += (cur - a[i].num) * a[i].t;
    }
    has.insert(cur);
    if (has.find(cur - 1) != has.end()) fa[cur - 1] = cur;
    if (has.find(cur + 1) != has.end()) fa[cur] = cur + 1;
  }
  cout << ans << '\n';
  return 0;
}
