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;
long long mod = 1000000000 + 7;
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
void fun() {}
bool isPrime(long long n) {
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
const long long N = 2e5 + 107;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fun();
  long long tt = 1;
  while (tt--) {
    long long n;
    cin >> n;
    long long a[n], t[n];
    long long h[N] = {0};
    for (long long i = 0; i < n; i++) cin >> a[i], h[a[i]]++;
    for (long long i = 0; i < n; i++) cin >> t[i];
    vector<pair<long long, long long> > v;
    for (long long i = 0; i < n; i++) v.push_back(make_pair(a[i], t[i]));
    sort(v.begin(), v.end());
    long long sum = 0, ans = 0, curr;
    priority_queue<long long> pq;
    for (long long i = 0; i < n; i++) {
      sum += v[i].second;
      pq.push(v[i].second);
      curr = v[i].first;
      while (not pq.empty() and (i == n - 1 or curr != v[i + 1].first)) {
        sum -= pq.top();
        pq.pop();
        ans += sum;
        curr++;
      }
    }
    cout << ans;
  }
  return 0;
}
