#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
struct Node {
  int a, t;
  inline Node() { a = t = 0; }
  inline Node(register int m_a, register int m_t) { a = m_a, t = m_t; }
  inline bool operator<(const register Node& b) const {
    if (a == b.a) return t > b.t;
    return a < b.a;
  }
} a[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  register int n;
  cin >> n;
  for (register int i = 1; i <= n; ++i) cin >> a[i].a;
  for (register int i = 1; i <= n; ++i) cin >> a[i].t;
  sort(a + 1, a + n + 1);
  register long long ans = 0;
  register priority_queue<int> que;
  register int i = 1;
  register int cur = a[1].a;
  register long long sum = 0;
  while (true) {
    if (que.empty() && i > n) break;
    while (i <= n && a[i].a == cur) {
      que.push(a[i].t);
      sum += a[i].t;
      ++i;
    }
    if (!que.empty()) {
      sum -= que.top();
      que.pop();
    }
    if (!que.empty()) {
      ++cur;
    } else
      cur = a[i].a;
    ans += sum;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int maxn = 2e5 + 10;
struct Node {
  int a, t;
  inline Node() { a = t = 0; }
  inline Node(register int m_a, register int m_t) { a = m_a, t = m_t; }
  inline bool operator<(const register Node& b) const {
    if (a == b.a) return t > b.t;
    return a < b.a;
  }
} a[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  register int n;
  cin >> n;
  for (register int i = 1; i <= n; ++i) cin >> a[i].a;
  for (register int i = 1; i <= n; ++i) cin >> a[i].t;
  sort(a + 1, a + n + 1);
  register long long ans = 0;
  register priority_queue<int> que;
  register int i = 1;
  register int cur = a[1].a;
  register long long sum = 0;
  while (true) {
    if (que.empty() && i > n) break;
    while (i <= n && a[i].a == cur) {
      que.push(a[i].t);
      sum += a[i].t;
      ++i;
    }
    if (!que.empty()) {
      sum -= que.top();
      que.pop();
    }
    if (!que.empty()) {
      ++cur;
    } else
      cur = a[i].a;
    ans += sum;
  }
  cout << ans << '\n';
  return 0;
}

```