#include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000000000000;
bool sortCct(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first > b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, i, t, g, c = 0;
  cin >> n;
  vector<long long> ct(n);
  vector<pair<long long, long long> > ctt(n);
  map<long long, long long> fr;
  for (i = 0; i < n; i++) cin >> ct[i];
  for (i = 0; i < n; i++) {
    cin >> t;
    ctt[i] = pair<long long, long long>(t, ct[i]);
  }
  sort(ctt.begin(), ctt.end(), sortCct);
  fr[INF] = 1;
  for (i = 0; i < n; i++) {
    pair<long long, long long> k = ctt[i];
    auto it = fr.lower_bound(k.second);
    g = max(k.second, it->second);
    c += (g - k.second) * k.first;
    if (g == it->second and g == it->first)
      fr.erase(it);
    else if (g == it->second)
      it->second++;
    else if (g == it->first) {
      fr[it->first - 1] = it->second;
      fr.erase(it);
    } else {
      fr[g - 1] = it->second;
      it->second = g + 1;
    }
  }
  cout << c << "\n";
  return 0;
}

### Prompt

Please create a solution in CPP 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 INF = 1000000000000000;
bool sortCct(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first > b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, i, t, g, c = 0;
  cin >> n;
  vector<long long> ct(n);
  vector<pair<long long, long long> > ctt(n);
  map<long long, long long> fr;
  for (i = 0; i < n; i++) cin >> ct[i];
  for (i = 0; i < n; i++) {
    cin >> t;
    ctt[i] = pair<long long, long long>(t, ct[i]);
  }
  sort(ctt.begin(), ctt.end(), sortCct);
  fr[INF] = 1;
  for (i = 0; i < n; i++) {
    pair<long long, long long> k = ctt[i];
    auto it = fr.lower_bound(k.second);
    g = max(k.second, it->second);
    c += (g - k.second) * k.first;
    if (g == it->second and g == it->first)
      fr.erase(it);
    else if (g == it->second)
      it->second++;
    else if (g == it->first) {
      fr[it->first - 1] = it->second;
      fr.erase(it);
    } else {
      fr[g - 1] = it->second;
      it->second = g + 1;
    }
  }
  cout << c << "\n";
  return 0;
}

```