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;
pair<long long, int> a[200005];
vector<int> v[200005];
long long hm[200005];
priority_queue<int> pq;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i].first);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].second);
  }
  sort(a + 1, a + n + 1);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i].first != a[i - 1].first) {
      v[++cnt].push_back(a[i].second);
      hm[cnt] = a[i].first;
    } else {
      v[cnt].push_back(a[i].second);
    }
  }
  long long ans = 0, sum = 0;
  for (int i = 1; i <= cnt; i++) {
    while (!pq.empty() && hm[i - 1] < hm[i]) {
      sum -= pq.top();
      pq.pop();
      ans += sum;
      hm[i - 1]++;
    }
    int sizee = v[i].size() - 1;
    for (int j = 0; j < sizee; j++) {
      sum += v[i][j];
      pq.push(v[i][j]);
    }
    if (pq.empty()) continue;
    if (v[i][sizee] >= pq.top()) {
      ans += sum;
    } else {
      sum -= pq.top();
      pq.pop();
      sum += v[i][sizee];
      pq.push(v[i][sizee]);
      ans += sum;
    }
    hm[i]++;
  }
  int num = 0;
  while (!pq.empty()) {
    ans += (long long)pq.top() * num;
    pq.pop();
    num++;
  }
  printf("%I64d\n", ans);
  return 0;
}
