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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const long long int mod = 1e9 + 7;
const long long int maxn = 5 * 100001;
long long int power(long long int x, long long int n) {
  long long int res = 1;
  for (; n > 0; n >>= 1) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  pair<long long int, long long int> a[n];
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i].first;
    m[a[i].first];
  }
  for (long long int i = 0; i < n; i++) {
    cin >> a[i].second;
  }
  long long int c = 0;
  map<long long int, long long int> rm;
  for (auto& x : m) {
    x.second = c++;
    rm[c - 1] = x.first;
  }
  rm[c] = 1e10;
  rm[-1] = rm[0];
  vector<long long int> g[m.size()];
  for (long long int i = 0; i < n; i++) {
    g[m[a[i].first]].push_back(a[i].second);
  }
  long long int ans = 0;
  long long int val = 0;
  multiset<long long int> curr;
  for (long long int i = 0; i < m.size(); i++) {
    ans += (rm[i] - rm[i - 1]) * val;
    for (auto x : g[i]) {
      val += x;
      curr.insert(x);
    }
    val -= *curr.rbegin();
    curr.erase(curr.find(*curr.rbegin()));
    c = 0;
    while (c < rm[i + 1] - rm[i] - 1 && curr.size() >= 1) {
      val -= *curr.rbegin();
      ans += (c + 1) * (*curr.rbegin());
      curr.erase(curr.find(*curr.rbegin()));
      c++;
    }
    if (i == m.size() - 1) assert(curr.size() == 0);
  }
  cout << ans << endl;
}
