#include <bits/stdc++.h>
using namespace std;
const char el = '\n';
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<long long> t(n);
  map<long long, vector<long long> > timeswithcount{};
  long long mincount = LLONG_MAX;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    mincount = min(mincount, a[i]);
  }
  for (int i = 0; i < n; ++i) {
    cin >> t[i];
    timeswithcount[a[i]].push_back(t[i]);
  }
  priority_queue<pair<long long, long long> > q;
  long long numassigned = 0;
  long long lastassigned = LLONG_MIN;
  long long currloc = mincount;
  long long ret = 0;
  while (numassigned < n) {
    if (timeswithcount.count(currloc)) {
      for (auto t : timeswithcount[currloc]) {
        q.push({t, currloc});
      }
      timeswithcount.erase(currloc);
    }
    if (q.empty()) {
      currloc = (*timeswithcount.begin()).first;
      continue;
    }
    auto mostexpensivepr = q.top();
    long long mostexpensiveT = mostexpensivepr.first;
    long long origloc = mostexpensivepr.second;
    ret += (mostexpensiveT) * (currloc - origloc);
    q.pop();
    ++currloc;
    ++numassigned;
  }
  cout << ret << el;
}

### Prompt

Please formulate 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 char el = '\n';
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<long long> t(n);
  map<long long, vector<long long> > timeswithcount{};
  long long mincount = LLONG_MAX;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    mincount = min(mincount, a[i]);
  }
  for (int i = 0; i < n; ++i) {
    cin >> t[i];
    timeswithcount[a[i]].push_back(t[i]);
  }
  priority_queue<pair<long long, long long> > q;
  long long numassigned = 0;
  long long lastassigned = LLONG_MIN;
  long long currloc = mincount;
  long long ret = 0;
  while (numassigned < n) {
    if (timeswithcount.count(currloc)) {
      for (auto t : timeswithcount[currloc]) {
        q.push({t, currloc});
      }
      timeswithcount.erase(currloc);
    }
    if (q.empty()) {
      currloc = (*timeswithcount.begin()).first;
      continue;
    }
    auto mostexpensivepr = q.top();
    long long mostexpensiveT = mostexpensivepr.first;
    long long origloc = mostexpensivepr.second;
    ret += (mostexpensiveT) * (currloc - origloc);
    q.pop();
    ++currloc;
    ++numassigned;
  }
  cout << ret << el;
}

```