#include <bits/stdc++.h>
using namespace std;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  vector<int> a(n), c(n);
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  for (int i = 0; i < (int)(n); ++i) cin >> c[i];
  vector<int> ord(n);
  iota((ord).begin(), (ord).end(), 0);
  sort((ord).begin(), (ord).end(), [&](int i, int j) { return a[i] < a[j]; });
  int L = 0, X = 0;
  multiset<int> cs;
  long long scs = 0, ans = 0;
  while (L < n || !cs.empty()) {
    if (cs.empty()) uax(X, a[ord[L]]);
    while (L < n && a[ord[L]] == X) {
      cs.insert(c[ord[L]]);
      scs += c[ord[L]];
      ++L;
    }
    if (!cs.empty()) {
      int z = *cs.rbegin();
      scs -= z;
      cs.erase(cs.find(z));
    }
    ans += scs;
    ++X;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  vector<int> a(n), c(n);
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  for (int i = 0; i < (int)(n); ++i) cin >> c[i];
  vector<int> ord(n);
  iota((ord).begin(), (ord).end(), 0);
  sort((ord).begin(), (ord).end(), [&](int i, int j) { return a[i] < a[j]; });
  int L = 0, X = 0;
  multiset<int> cs;
  long long scs = 0, ans = 0;
  while (L < n || !cs.empty()) {
    if (cs.empty()) uax(X, a[ord[L]]);
    while (L < n && a[ord[L]] == X) {
      cs.insert(c[ord[L]]);
      scs += c[ord[L]];
      ++L;
    }
    if (!cs.empty()) {
      int z = *cs.rbegin();
      scs -= z;
      cs.erase(cs.find(z));
    }
    ans += scs;
    ++X;
  }
  cout << ans << '\n';
  return 0;
}

```