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;
long long p = 1e9 + 7;
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % p;
    }
    a = (a * a) % p;
    b /= 2;
  }
  return res;
}
long long inv(long long a) { return power(a, p - 2); }
void dijks(vector<vector<pair<int, int> > > &adj, int s, vector<int> &d) {
  d[s] = 0;
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      pq;
  pq.push(make_pair(d[s], 0));
  while (!pq.empty()) {
    pair<int, int> temp = pq.top();
    pq.pop();
    int u = temp.second;
    if (d[u] < temp.first) continue;
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i].first;
      int wt = adj[u][i].second;
      if (d[v] <= d[u] + wt)
        continue;
      else {
        d[v] = d[u] + wt;
        pq.push(make_pair(d[v], v));
      }
    }
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
int main() {
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<long long, long long> > a(n);
    for (int i = 0; i < n; i++) cin >> a[i].first;
    for (int i = 0; i < n; i++) cin >> a[i].second;
    long long ans = 0;
    sort(a.begin(), a.end());
    priority_queue<long long> pq;
    long long sum = 0;
    for (int i = 0; i < n;) {
      long long ind = i;
      long long num = a[i].first;
      while (num == a[i].first) i++;
      if (!pq.empty() || i - ind > 1) {
        for (int x = ind; x < i; x++) {
          sum += a[x].second;
          pq.push(a[x].second);
        }
        while (i == n || num < a[i].first) {
          if (!pq.empty())
            sum -= pq.top();
          else
            break;
          pq.pop();
          ans += sum;
          num++;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
