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;
void setup() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
}
signed main() {
  setup();
  long long n;
  cin >> n;
  map<long long, vector<long long>> a;
  vector<pair<long long, long long>> tmp(n);
  for (long long i = 0; i < n; ++i) cin >> tmp[i].first;
  map<long long, long long> sum;
  for (long long i = 0; i < n; ++i) cin >> tmp[i].second;
  for (auto i : tmp) {
    sum[i.first] = 0;
    a[i.first].push_back(i.second);
  }
  long long ans = 0;
  multiset<long long> now;
  for (auto i : sum) {
    for (auto j : a[i.first]) {
      i.second += j;
      now.insert(j);
    }
    ans += i.second - *now.rbegin();
    if (i.second - *now.rbegin()) sum[i.first + 1] += i.second - *now.rbegin();
    now.erase(now.find(*now.rbegin()));
  }
  cout << ans << '\n';
}
