#include <bits/stdc++.h>
using namespace std;
ifstream fin("date.in");
struct data {
  int val;
  int poz;
} adaugare;
bool operator<(data a, data b) { return a.val < b.val; }
multiset<data> s;
multiset<data>::iterator its;
int w[200005];
map<long long, queue<int> > v;
map<long long, queue<int> >::iterator it1, it2;
int main() {
  long long i, j, n, x, poz1, poz2, dif, ct, poz_curenta, poz_adaugat, aux;
  long long suma = 0;
  cin >> n;
  for (i = 0; i < n; i++) cin >> w[i];
  for (i = 0; i < n; i++) {
    cin >> x;
    v[w[i]].push(x);
  }
  for (it1 = v.begin(), it2 = v.begin(), it2++; it2 != v.end(); it1++, it2++) {
    poz1 = (*it1).first;
    poz2 = (*it2).first;
    dif = poz2 - poz1;
    while (!(*it1).second.empty()) {
      adaugare.val = (*it1).second.front();
      adaugare.poz = poz1;
      s.insert(adaugare);
      (*it1).second.pop();
    }
    for (i = 0, its = s.end(), its--; i < dif - 1 && its != s.begin(); i++) {
      poz_curenta = poz1 + i;
      poz_adaugat = (*its).poz;
      aux = (*its).val;
      suma += (*its).val * (poz_curenta - poz_adaugat);
      its = s.erase(its);
      its--;
    }
    poz_curenta = poz1 + i;
    poz_adaugat = (*its).poz;
    suma += (*its).val * (poz_curenta - poz_adaugat);
    its = s.erase(its);
  }
  poz1 = (*it1).first;
  while (!(*it1).second.empty()) {
    adaugare.val = (*it1).second.front();
    adaugare.poz = poz1;
    s.insert(adaugare);
    (*it1).second.pop();
  }
  for (i = 0, its = s.end(), its--; its != s.begin(); i++) {
    poz_curenta = poz1 + i;
    poz_adaugat = (*its).poz;
    aux = (*its).val;
    suma += (*its).val * (poz_curenta - poz_adaugat);
    its = s.erase(its);
    its--;
  }
  poz_curenta = poz1 + i;
  poz_adaugat = (*its).poz;
  suma += (*its).val * (poz_curenta - poz_adaugat);
  cout << suma;
  return 0;
}

### Prompt

Your challenge is to write 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;
ifstream fin("date.in");
struct data {
  int val;
  int poz;
} adaugare;
bool operator<(data a, data b) { return a.val < b.val; }
multiset<data> s;
multiset<data>::iterator its;
int w[200005];
map<long long, queue<int> > v;
map<long long, queue<int> >::iterator it1, it2;
int main() {
  long long i, j, n, x, poz1, poz2, dif, ct, poz_curenta, poz_adaugat, aux;
  long long suma = 0;
  cin >> n;
  for (i = 0; i < n; i++) cin >> w[i];
  for (i = 0; i < n; i++) {
    cin >> x;
    v[w[i]].push(x);
  }
  for (it1 = v.begin(), it2 = v.begin(), it2++; it2 != v.end(); it1++, it2++) {
    poz1 = (*it1).first;
    poz2 = (*it2).first;
    dif = poz2 - poz1;
    while (!(*it1).second.empty()) {
      adaugare.val = (*it1).second.front();
      adaugare.poz = poz1;
      s.insert(adaugare);
      (*it1).second.pop();
    }
    for (i = 0, its = s.end(), its--; i < dif - 1 && its != s.begin(); i++) {
      poz_curenta = poz1 + i;
      poz_adaugat = (*its).poz;
      aux = (*its).val;
      suma += (*its).val * (poz_curenta - poz_adaugat);
      its = s.erase(its);
      its--;
    }
    poz_curenta = poz1 + i;
    poz_adaugat = (*its).poz;
    suma += (*its).val * (poz_curenta - poz_adaugat);
    its = s.erase(its);
  }
  poz1 = (*it1).first;
  while (!(*it1).second.empty()) {
    adaugare.val = (*it1).second.front();
    adaugare.poz = poz1;
    s.insert(adaugare);
    (*it1).second.pop();
  }
  for (i = 0, its = s.end(), its--; its != s.begin(); i++) {
    poz_curenta = poz1 + i;
    poz_adaugat = (*its).poz;
    aux = (*its).val;
    suma += (*its).val * (poz_curenta - poz_adaugat);
    its = s.erase(its);
    its--;
  }
  poz_curenta = poz1 + i;
  poz_adaugat = (*its).poz;
  suma += (*its).val * (poz_curenta - poz_adaugat);
  cout << suma;
  return 0;
}

```