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;
const int MAX = (2e5) + 10;
pair<int, int> ara[MAX];
int id[MAX];
vector<int> didi[MAX];
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i <= n - 1; i++) {
    cin >> ara[i].first;
  }
  for (int i = 0; i <= n - 1; i++) {
    cin >> ara[i].second;
  }
  sort(ara, ara + n, [](pair<int, int> x, pair<int, int> y) {
    return (x.first == y.first) ? x.second > y.second : x.first < y.first;
  });
  int cp = -1;
  priority_queue<int> que;
  long long sum, prin;
  sum = prin = 0;
  for (int i = 0; i <= n - 1; i++) {
    if (cp >= ara[i].first) {
      que.push(ara[i].second);
      sum += ara[i].second;
    } else {
      if (!que.empty()) {
        sum -= que.top();
        que.pop();
        prin += sum;
        cp++;
        i--;
      } else {
        cp = ara[i].first;
        i--;
      }
    }
  }
  while (!que.empty()) {
    sum -= que.top();
    que.pop();
    prin += sum;
  }
  cout << prin << endl;
}
int main() {
  ios::sync_with_stdio(false);
  int tc;
  solve();
  return 0;
}
