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>
#pragma GCC optimize("O3")
using namespace std;
long long n;
long long cost[((long long)2e5 + 5)];
long long ans = 0;
vector<pair<long long, long long>> v;
map<long long, long long> used;
map<long long, set<long long>> s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  v.resize(n);
  for (long long i = 0; i < n; ++i) {
    cin >> v[i].first;
  }
  for (long long i = 0; i < n; ++i) cin >> v[i].second;
  sort(v.begin(), v.end());
  priority_queue<long long> q;
  long long weight = 0;
  long long t = v[0].second;
  long long it = 0;
  while (v[it].first == t) {
    q.push({v[it].second});
    weight += v[it].second;
    it++;
  }
  long long num = t;
  while (1) {
    while (q.empty() == false) {
      weight -= q.top();
      q.pop();
      ans += weight;
      num++;
      while (it < v.size() && v[it].first == num) {
        q.push({v[it].second});
        weight += v[it].second;
        it++;
      }
    }
    if (it == v.size()) break;
    num = v[it].first;
    while (it < v.size() && v[it].first == num) {
      q.push({v[it].second});
      weight += v[it].second;
      it++;
    }
  }
  cout << ans << "\n";
}
