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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
const int INF = 1 << 29;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
long long int a[200005], cost[200005];
int main() {
  multiset<long long int> num;
  map<long long int, vector<long long int> > mp;
  set<long long int> indices;
  long long int n;
  long long int ans = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    indices.insert(a[i]);
  }
  for (int i = 0; i < n; i++) {
    cin >> cost[i];
    mp[a[i]].push_back(cost[i]);
  }
  long long int ind = *indices.begin();
  long long int sum = 0;
  while (!num.empty() || ind <= *indices.rbegin()) {
    if (mp.find(ind) != mp.end())
      for (auto i = mp[ind].begin(); i != mp[ind].end(); i++)
        num.insert(*i), sum += *i;
    sum -= *num.rbegin();
    ans += sum;
    auto k = num.end();
    k--;
    num.erase(k);
    if (num.empty() && ind < *indices.rbegin()) {
      ind = *indices.upper_bound(ind);
      continue;
    }
    ind++;
  }
  cout << ans;
}
