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 INF = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const long long mod = 1e9 + 7;
map<long long, long long> m;
vector<pair<long long, long long> > num;
long long find(long long x) {
  if (m[x] == 0) return x;
  return m[x] = find(m[x]);
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return a > b;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    pair<long long, long long> p;
    cin >> p.second;
    num.push_back(p);
  }
  for (pair<long long, long long> &p : num) {
    cin >> p.first;
  }
  sort(num.begin(), num.end(), cmp);
  long long ans = 0;
  for (pair<long long, long long> p : num) {
    long long x = find(p.second);
    m[x] = x + 1;
    ans += (x - p.second) * p.first;
  }
  cout << ans << '\n';
  return 0;
}
