#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long po(long long x, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * x);
    x = (x * x);
    n /= 2;
  }
  return ans;
}
void func() {
  long long n;
  cin >> n;
  map<long long, vector<long long> > m;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    long long t;
    cin >> t;
    m[a[i]].push_back(t);
  }
  long long i = 0;
  long long ans = 0;
  while (1) {
    auto j = m.upper_bound(i);
    if (j == m.end()) break;
    long long k = j->first;
    multiset<long long> temp;
    long long sum = 0;
    for (int x = 0; x < m[k].size(); x++) {
      temp.insert(m[k][x]);
      sum += m[k][x];
    }
    while (1) {
      if (temp.size() == 1) break;
      sum -= *(temp.rbegin());
      auto z = temp.end();
      z--;
      temp.erase(z);
      k++;
      ans += sum;
      for (int x = 0; x < m[k].size(); x++) {
        temp.insert(m[k][x]);
        sum += m[k][x];
      }
    }
    i = k;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) func();
}

### Prompt

Generate a CPP solution 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;
const long long mod = 1000000007;
long long po(long long x, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * x);
    x = (x * x);
    n /= 2;
  }
  return ans;
}
void func() {
  long long n;
  cin >> n;
  map<long long, vector<long long> > m;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    long long t;
    cin >> t;
    m[a[i]].push_back(t);
  }
  long long i = 0;
  long long ans = 0;
  while (1) {
    auto j = m.upper_bound(i);
    if (j == m.end()) break;
    long long k = j->first;
    multiset<long long> temp;
    long long sum = 0;
    for (int x = 0; x < m[k].size(); x++) {
      temp.insert(m[k][x]);
      sum += m[k][x];
    }
    while (1) {
      if (temp.size() == 1) break;
      sum -= *(temp.rbegin());
      auto z = temp.end();
      z--;
      temp.erase(z);
      k++;
      ans += sum;
      for (int x = 0; x < m[k].size(); x++) {
        temp.insert(m[k][x]);
        sum += m[k][x];
      }
    }
    i = k;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) func();
}

```