#include <bits/stdc++.h>
template <typename Tp>
void read(Tp &res) {
  static char ch;
  ch = getchar(), res = 0;
  static bool flag;
  flag = 0;
  while (!isdigit(ch)) flag = (ch == '-'), ch = getchar();
  while (isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
  if (flag) res = -res;
}
const int maxn = 2e5 + 19;
long long ans;
int n;
std::pair<int, int> a[maxn];
std::priority_queue<std::pair<int, int>> q;
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(a[i].first);
  for (int i = 1; i <= n; ++i) read(a[i].second);
  std::sort(a + 1, a + 1 + n), a[n + 1].first = 2e9;
  for (int i = 1; i <= n; ++i) {
    int r = i;
    q.emplace(a[r].second, a[r].first);
    while (r < n && a[r + 1].first == a[r].first)
      ++r, q.emplace(a[r].second, a[r].first);
    i = r;
    for (int j = a[i].first; j < a[i + 1].first && q.size(); ++j)
      ans += (long long)q.top().first * (j - q.top().second), q.pop();
  }
  printf("%lld\n", ans);
}

### 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>
template <typename Tp>
void read(Tp &res) {
  static char ch;
  ch = getchar(), res = 0;
  static bool flag;
  flag = 0;
  while (!isdigit(ch)) flag = (ch == '-'), ch = getchar();
  while (isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
  if (flag) res = -res;
}
const int maxn = 2e5 + 19;
long long ans;
int n;
std::pair<int, int> a[maxn];
std::priority_queue<std::pair<int, int>> q;
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(a[i].first);
  for (int i = 1; i <= n; ++i) read(a[i].second);
  std::sort(a + 1, a + 1 + n), a[n + 1].first = 2e9;
  for (int i = 1; i <= n; ++i) {
    int r = i;
    q.emplace(a[r].second, a[r].first);
    while (r < n && a[r + 1].first == a[r].first)
      ++r, q.emplace(a[r].second, a[r].first);
    i = r;
    for (int j = a[i].first; j < a[i + 1].first && q.size(); ++j)
      ans += (long long)q.top().first * (j - q.top().second), q.pop();
  }
  printf("%lld\n", ans);
}

```