#include <bits/stdc++.h>
using namespace std;
long long modd = 1000 * 1000 * 1000 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cout.precision(17);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  uniform_int_distribution<int> rand_gen(0, 1000);
  int test_cases = 1;
  for (int t = (1); t < (1) + (test_cases); ++t) {
    int n;
    cin >> n;
    vector<long long> a, time_needed;
    for (int i = (0); i < (0) + (n); ++i) {
      int kk;
      cin >> kk;
      a.push_back(kk);
    }
    for (int i = (0); i < (0) + (n); ++i) {
      int kk;
      cin >> kk;
      time_needed.push_back(kk);
    }
    map<long long, multiset<long long>> freq;
    for (int i = (0); i < (0) + (n); ++i) {
      freq[a[i]].insert(time_needed[i]);
    }
    long long ret = 0;
    multiset<pair<long long, long long>> curr;
    long long val = freq.begin()->first;
    while (true) {
      for (auto x : freq[val]) {
        curr.insert(make_pair(x, val));
      }
      if (!curr.empty()) {
        ret += (val - prev(curr.end())->second) * prev(curr.end())->first;
        curr.erase(prev(curr.end()));
      }
      ++val;
      if (curr.empty() && (val <= prev(freq.end())->first)) {
        val = (freq.lower_bound(val))->first;
      }
      if ((val > prev(freq.end())->first) && (curr.empty())) {
        break;
      }
    }
    cout << ret << endl;
  }
  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;
long long modd = 1000 * 1000 * 1000 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cout.precision(17);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  uniform_int_distribution<int> rand_gen(0, 1000);
  int test_cases = 1;
  for (int t = (1); t < (1) + (test_cases); ++t) {
    int n;
    cin >> n;
    vector<long long> a, time_needed;
    for (int i = (0); i < (0) + (n); ++i) {
      int kk;
      cin >> kk;
      a.push_back(kk);
    }
    for (int i = (0); i < (0) + (n); ++i) {
      int kk;
      cin >> kk;
      time_needed.push_back(kk);
    }
    map<long long, multiset<long long>> freq;
    for (int i = (0); i < (0) + (n); ++i) {
      freq[a[i]].insert(time_needed[i]);
    }
    long long ret = 0;
    multiset<pair<long long, long long>> curr;
    long long val = freq.begin()->first;
    while (true) {
      for (auto x : freq[val]) {
        curr.insert(make_pair(x, val));
      }
      if (!curr.empty()) {
        ret += (val - prev(curr.end())->second) * prev(curr.end())->first;
        curr.erase(prev(curr.end()));
      }
      ++val;
      if (curr.empty() && (val <= prev(freq.end())->first)) {
        val = (freq.lower_bound(val))->first;
      }
      if ((val > prev(freq.end())->first) && (curr.empty())) {
        break;
      }
    }
    cout << ret << endl;
  }
  return 0;
}

```