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;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '[';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << ']';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '{' << p.first << ", " << p.second << '}';
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &s) {
  os << '{';
  string sep;
  for (const auto &x : s) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const map<A, B> &mp) {
  os << '[';
  string sep;
  for (const auto &x : mp) os << sep << x, sep = ", ";
  return os << ']';
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &s) {
  os << '{';
  string sep;
  for (const auto &x : s) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
void solve() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> v(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());
  long long ans = 0;
  priority_queue<pair<long long, long long>> pq;
  for (long long i = 0; i < n;) {
    long long j = 0;
    for (j = i; j < n && (v[i].first == v[j].first); j++) {
      pq.push({v[j].second, v[j].first});
    }
    long long last_limit = (j == n ? 1e18 : v[j].first);
    long long now = v[i].first;
    i = j;
    while ((long long)(pq).size() && (now < last_limit)) {
      pair<long long, long long> here = pq.top();
      pq.pop();
      ans += (now - here.second) * here.first;
      now++;
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  solve();
  return 0;
}
