#include <bits/stdc++.h>
using namespace std;
long long n, h, l, r;
long long fun(long long i, long long t, vector<long long> &a,
              vector<vector<long long>> &dp) {
  if (i >= n) return 0;
  t %= h;
  if (dp[i][t] != -1) return dp[i][t];
  long long m = (t >= l and t <= r);
  m +=
      max(fun(i + 1, t + a[i + 1], a, dp), fun(i + 1, t + a[i + 1] - 1, a, dp));
  return dp[i][t] = m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = 0;
  }
  for (long long i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long ans = 0;
  multiset<long long> ms;
  long long i = 0, c = 0, s = 0;
  while (i < n) {
    if (ms.empty()) c = a[i].first;
    while (i < n and a[i].first == c) {
      ms.insert(a[i].second);
      s += a[i].second;
      i++;
    }
    while (ms.size()) {
      if (i < n and a[i].first == c) break;
      auto it = ms.end();
      it--;
      s -= *it;
      ans += s;
      ms.erase(it);
      c++;
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long long n, h, l, r;
long long fun(long long i, long long t, vector<long long> &a,
              vector<vector<long long>> &dp) {
  if (i >= n) return 0;
  t %= h;
  if (dp[i][t] != -1) return dp[i][t];
  long long m = (t >= l and t <= r);
  m +=
      max(fun(i + 1, t + a[i + 1], a, dp), fun(i + 1, t + a[i + 1] - 1, a, dp));
  return dp[i][t] = m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = 0;
  }
  for (long long i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  long long ans = 0;
  multiset<long long> ms;
  long long i = 0, c = 0, s = 0;
  while (i < n) {
    if (ms.empty()) c = a[i].first;
    while (i < n and a[i].first == c) {
      ms.insert(a[i].second);
      s += a[i].second;
      i++;
    }
    while (ms.size()) {
      if (i < n and a[i].first == c) break;
      auto it = ms.end();
      it--;
      s -= *it;
      ans += s;
      ms.erase(it);
      c++;
    }
  }
  cout << ans << '\n';
  return 0;
}

```