#include <bits/stdc++.h>
using namespace std;
priority_queue<long long, vector<long long>, less<long long> > q;
const long long N = 200005;
struct node {
  long long a, t;
} op[N];
long long n, a[N], t[N], sum = 0, ans = 0;
bool oppp(node R, node T) {
  if (R.a == T.a) return R.t > T.t;
  return R.a < T.a;
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &op[i].a);
  for (long long i = 1; i <= n; i++) scanf("%lld", &op[i].t);
  sort(op + 1, op + 1 + n, oppp);
  op[n + 1].a = 999999999999;
  for (long long i = 2; i <= n; i++) {
    if (op[i].a == op[i - 1].a) {
      q.push(op[i].t);
      sum += op[i].t;
    }
    if (op[i].a != op[i - 1].a && !q.empty()) {
      {
        q.push(op[i].t);
        sum += op[i].t;
      }
      sum -= q.top();
      q.pop();
    }
    if (q.empty())
      continue;
    else {
      if ((op[i + 1].a - op[i].a - 1) >= q.size()) {
        long long qw = 1;
        while (!q.empty()) {
          sum -= q.top();
          ans += q.top() * qw;
          qw++;
          q.pop();
        }
      } else {
        for (long long j = 1; j <= (op[i + 1].a - op[i].a - 1); j++) {
          sum -= q.top();
          ans += q.top() * j;
          q.pop();
        }
        ans += (op[i + 1].a - op[i].a) * sum;
      }
    }
  }
  cout << ans << endl;
}

### Prompt

Please create a solution in CPP to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
priority_queue<long long, vector<long long>, less<long long> > q;
const long long N = 200005;
struct node {
  long long a, t;
} op[N];
long long n, a[N], t[N], sum = 0, ans = 0;
bool oppp(node R, node T) {
  if (R.a == T.a) return R.t > T.t;
  return R.a < T.a;
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &op[i].a);
  for (long long i = 1; i <= n; i++) scanf("%lld", &op[i].t);
  sort(op + 1, op + 1 + n, oppp);
  op[n + 1].a = 999999999999;
  for (long long i = 2; i <= n; i++) {
    if (op[i].a == op[i - 1].a) {
      q.push(op[i].t);
      sum += op[i].t;
    }
    if (op[i].a != op[i - 1].a && !q.empty()) {
      {
        q.push(op[i].t);
        sum += op[i].t;
      }
      sum -= q.top();
      q.pop();
    }
    if (q.empty())
      continue;
    else {
      if ((op[i + 1].a - op[i].a - 1) >= q.size()) {
        long long qw = 1;
        while (!q.empty()) {
          sum -= q.top();
          ans += q.top() * qw;
          qw++;
          q.pop();
        }
      } else {
        for (long long j = 1; j <= (op[i + 1].a - op[i].a - 1); j++) {
          sum -= q.top();
          ans += q.top() * j;
          q.pop();
        }
        ans += (op[i + 1].a - op[i].a) * sum;
      }
    }
  }
  cout << ans << endl;
}

```