#include <bits/stdc++.h>
using namespace std;
const long long MAXN = (int)3e5 + 5;
long gcd(long a, long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  map<long long, long long> m, seon;
  long long a[n], t[n];
  set<long long> s;
  for (long int i = (long int)0; i < (long int)n; i++) {
    cin >> a[i];
    seon[a[i]]++;
  }
  for (auto i : seon) {
    long long val = i.second - 1, ind = i.first;
    if (val >= 0) s.insert(i.first);
    if (val > 0) {
      ind++;
      while (seon[ind] == 0 && val > 0) {
        seon[ind] = 1;
        s.insert(ind);
        ind++;
        val--;
      }
      seon[ind] += val;
    }
  }
  vector<pair<long, long> > v;
  for (long int i = (long int)0; i < (long int)n; i++) {
    cin >> t[i];
    v.push_back({t[i], i});
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  long long ans = 0;
  for (long int i = (long int)0; i < (long int)n; i++) {
    long long ind = a[v[i].second];
    if (m[ind] == 0) {
      m[ind] = 1;
      s.erase(ind);
    } else {
      long long num = *s.upper_bound(ind);
      ans += (num - ind) * v[i].first;
      m[num] = 1;
      s.erase(num);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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 MAXN = (int)3e5 + 5;
long gcd(long a, long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  map<long long, long long> m, seon;
  long long a[n], t[n];
  set<long long> s;
  for (long int i = (long int)0; i < (long int)n; i++) {
    cin >> a[i];
    seon[a[i]]++;
  }
  for (auto i : seon) {
    long long val = i.second - 1, ind = i.first;
    if (val >= 0) s.insert(i.first);
    if (val > 0) {
      ind++;
      while (seon[ind] == 0 && val > 0) {
        seon[ind] = 1;
        s.insert(ind);
        ind++;
        val--;
      }
      seon[ind] += val;
    }
  }
  vector<pair<long, long> > v;
  for (long int i = (long int)0; i < (long int)n; i++) {
    cin >> t[i];
    v.push_back({t[i], i});
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  long long ans = 0;
  for (long int i = (long int)0; i < (long int)n; i++) {
    long long ind = a[v[i].second];
    if (m[ind] == 0) {
      m[ind] = 1;
      s.erase(ind);
    } else {
      long long num = *s.upper_bound(ind);
      ans += (num - ind) * v[i].first;
      m[num] = 1;
      s.erase(num);
    }
  }
  cout << ans << endl;
  return 0;
}

```