#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long longinf = 1LL << 60;
const long long mod = 1e9 + 7;
struct cat {
  long long a;
  int t;
  cat() : a(0), t(0) {}
};
bool comp(cat c1, cat c2) { return c1.a < c2.a; }
int main() {
  int N;
  cin >> N;
  vector<cat> C(N);
  for (int i = (int)(0); i < (int)(N); i++) cin >> C[i].a;
  for (int i = (int)(0); i < (int)(N); i++) cin >> C[i].t;
  sort(C.begin(), C.end(), comp);
  priority_queue<int> q;
  int i = 0;
  long long ans = 0;
  long long sum = 0;
  int prev = 0;
  while (i < N) {
    int tmp = C[i].a;
    while (C[i].a == tmp) {
      q.push(C[i].t);
      sum += C[i].t;
      i++;
    }
    while (!q.empty() && i < N && tmp < C[i].a) {
      int top = q.top();
      sum -= top;
      q.pop();
      ans += sum;
      tmp++;
    }
  }
  while (!q.empty()) {
    int top = q.top();
    sum -= top;
    q.pop();
    ans += sum;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int inf = 1e9 + 7;
const long long longinf = 1LL << 60;
const long long mod = 1e9 + 7;
struct cat {
  long long a;
  int t;
  cat() : a(0), t(0) {}
};
bool comp(cat c1, cat c2) { return c1.a < c2.a; }
int main() {
  int N;
  cin >> N;
  vector<cat> C(N);
  for (int i = (int)(0); i < (int)(N); i++) cin >> C[i].a;
  for (int i = (int)(0); i < (int)(N); i++) cin >> C[i].t;
  sort(C.begin(), C.end(), comp);
  priority_queue<int> q;
  int i = 0;
  long long ans = 0;
  long long sum = 0;
  int prev = 0;
  while (i < N) {
    int tmp = C[i].a;
    while (C[i].a == tmp) {
      q.push(C[i].t);
      sum += C[i].t;
      i++;
    }
    while (!q.empty() && i < N && tmp < C[i].a) {
      int top = q.top();
      sum -= top;
      q.pop();
      ans += sum;
      tmp++;
    }
  }
  while (!q.empty()) {
    int top = q.top();
    sum -= top;
    q.pop();
    ans += sum;
  }
  cout << ans << endl;
  return 0;
}

```