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 long double PI = acos(-1);
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  gcd(b, a % b);
}
void solve() {
  long long n;
  cin >> n;
  long long a[n + 1], t[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  vector<pair<long long, long long> > v;
  for (long long i = 1; i <= n; i++) {
    cin >> t[i];
    v.push_back({a[i], -t[i]});
  }
  sort(v.begin(), v.end());
  priority_queue<long long> q;
  for (long long i = 0; i < n; i++) {
    v[i].second *= -1;
  }
  long long sum = 0, ans = 0, d = 200005;
  for (long long i = 1; i < n; i++) {
    if (v[i].first == v[i - 1].first) {
      long long temp = v[i].second;
      q.push(temp);
      ans += temp;
      sum += temp;
    } else {
      if (v[i].first > v[i - 1].first) {
        long long diff = v[i].first - v[i - 1].first;
        diff -= 1;
        while (diff && q.size()) {
          long long temp = q.top();
          q.pop();
          sum -= temp;
          ans += sum;
          diff--;
        }
        q.push(v[i].second);
        sum += v[i].second;
        diff = 1;
        while (diff && q.size()) {
          long long temp = q.top();
          q.pop();
          sum -= temp;
          ans += sum;
          diff--;
        }
      }
    }
  }
  while (d && q.size()) {
    long long temp = q.top();
    q.pop();
    sum -= temp;
    ans += sum;
    d--;
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
