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 <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n), t(n);
  for (auto &i : a) cin >> i;
  for (auto &i : t) cin >> i;
  map<long long, vector<long long> > mp;
  for (long long i = (0); i < (n); i++) mp[a[i]].push_back(t[i]);
  vector<long long> vals;
  for (auto &i : mp) vals.push_back(i.first);
  multiset<long long> ms;
  long long sum = 0;
  long long ans = 0;
  long long curr = vals[0];
  long long ind = 1;
  while (true) {
    vector<long long> x = mp[curr];
    for (auto &i : x) {
      ms.insert(i);
      sum += i;
    }
    assert(ms.size() > 0);
    sum -= *ms.rbegin();
    ans += sum;
    ms.erase(ms.lower_bound(*ms.rbegin()));
    curr++;
    if (ms.size() == 0) {
      auto loc = lower_bound(vals.begin(), vals.end(), curr);
      if (loc == vals.end()) break;
      long long index = loc - vals.begin();
      curr = vals[index];
      ind++;
    }
  }
  cout << ans << '\n';
}
signed main() {
  cin.tie(NULL), ios_base::sync_with_stdio(false), cout.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}
