#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
void run() {
  long long n;
  cin >> n;
  vector<long long> a(n), cost(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; ++i) {
    cin >> cost[i];
  }
  vector<pair<long long, long long> > v(n);
  for (long long i = 0; i < n; ++i) {
    v[i] = {a[i], cost[i]};
  }
  sort(v.begin(), v.end());
  long long ans = 0;
  long long sum = 0;
  multiset<long long> costs;
  long long itr = 0;
  long long cur_val = v[0].first;
  while (itr < n || costs.size()) {
    if (costs.size() == 0 && v[itr].first != cur_val) {
      cur_val = v[itr].first;
    }
    while (itr < n && v[itr].first == cur_val) {
      costs.insert(v[itr].second);
      sum += v[itr].second;
      ++itr;
    }
    long long max_val = *costs.rbegin();
    ans += sum - max_val;
    costs.erase(costs.find(max_val));
    sum -= max_val;
    cur_val++;
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  run();
  return 0;
}

### Prompt

Your challenge is to write 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 = 1e9 + 7;
const long long INF = 1e18;
void run() {
  long long n;
  cin >> n;
  vector<long long> a(n), cost(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; ++i) {
    cin >> cost[i];
  }
  vector<pair<long long, long long> > v(n);
  for (long long i = 0; i < n; ++i) {
    v[i] = {a[i], cost[i]};
  }
  sort(v.begin(), v.end());
  long long ans = 0;
  long long sum = 0;
  multiset<long long> costs;
  long long itr = 0;
  long long cur_val = v[0].first;
  while (itr < n || costs.size()) {
    if (costs.size() == 0 && v[itr].first != cur_val) {
      cur_val = v[itr].first;
    }
    while (itr < n && v[itr].first == cur_val) {
      costs.insert(v[itr].second);
      sum += v[itr].second;
      ++itr;
    }
    long long max_val = *costs.rbegin();
    ans += sum - max_val;
    costs.erase(costs.find(max_val));
    sum -= max_val;
    cur_val++;
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  run();
  return 0;
}

```