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("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
using namespace std;
template <class T>
ostream &operator<<(ostream &str, vector<T> &a) {
  for (auto &i : a) {
    str << i << " ";
  }
  return str;
}
template <class T>
istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) {
    str >> i;
  }
  return str;
}
template <class T>
ostream &operator<<(ostream &str, pair<T, T> &a) {
  str << a.first << " " << a.second;
  return str;
}
template <class T>
istream &operator>>(istream &str, pair<T, T> &a) {
  str >> a.first >> a.second;
  return str;
}
void solve();
signed main() {
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.precision(9);
  solve();
  return 0;
}
const long long MOD = 1e9 + 7, MAXN = 1e6 + 10, INF = 1e18 + 10;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<long long> t(n);
  for (long long i = 0; i < n; ++i) {
    cin >> t[i];
  }
  map<long long, multiset<long long> *> mp;
  map<long long, long long> ss;
  for (long long i = 0; i < n; ++i) {
    mp[a[i]] = new multiset<long long>;
  }
  for (long long i = 0; i < n; ++i) {
    mp[a[i]]->insert(t[i]);
    ss[a[i]] += t[i];
  }
  mp[INF] = new multiset<long long>;
  multiset<long long> *p = mp.begin()->second;
  long long sum = 0;
  for (auto it : *p) {
    sum += it;
  }
  long long add = 0, pnum, ans = 0;
  for (auto it : mp) {
    if (it.first == mp.begin()->first) {
      pnum = it.first;
      continue;
    }
    long long num = it.first;
    long long dif = num - pnum;
    while (dif > 0 && p->size() > 0) {
      ans += (sum - *p->rbegin());
      sum -= *p->rbegin();
      p->erase(--p->end());
      dif--;
    }
    if (p->size() > it.second->size()) {
      for (auto it : *it.second) {
        p->insert(it);
        sum += it;
      }
    } else {
      sum += ss[num];
      for (auto itt : *p) {
        it.second->insert(itt);
      }
      p = it.second;
    }
    pnum = num;
  }
  cout << ans << '\n';
}
