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 MSIZE = 2e5 + 5;
const int MOD = 1e9 + 7;
map<int, vector<int>> mp;
int cost[MSIZE], arr[MSIZE];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  priority_queue<pair<long long, int>> pq;
  int n;
  cin >> n;
  for (int i = 0; i <= n - 1; i++) {
    cin >> arr[i];
    mp[arr[i]].push_back(i);
  }
  for (int i = 0; i <= n - 1; i++) cin >> cost[i];
  sort(arr, arr + n);
  long long ans = 0;
  for (int i = 0; i <= n - 1; i++) {
    if (i != 0 && arr[i - 1] == arr[i]) continue;
    for (int j : mp[arr[i]]) {
      pq.push({cost[j], arr[i]});
    }
    ans += pq.top().first * (arr[i] - pq.top().second);
    pq.pop();
    int nxt = arr[i] + 1;
    while (!pq.empty() && mp[nxt].empty()) {
      ans += pq.top().first * (nxt - pq.top().second);
      pq.pop();
      nxt++;
    }
  }
  cout << ans << endl;
}
