#include <bits/stdc++.h>
using namespace std;
const long double pi = acos(-1.0);
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
bool chmin(T &x, T y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
bool chmax(T &x, T y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
template <typename U, typename V>
ostream &operator<<(ostream &s, const pair<U, V> &x) {
  s << "(" << x.first << ", " << x.second << ")";
  return s;
}
template <typename U>
ostream &operator<<(ostream &s, const vector<U> &x) {
  s << "[";
  bool was = false;
  for (auto it : x) {
    if (was) {
      s << ", ";
    }
    was = true;
    s << it;
  }
  s << "]";
  return s;
}
template <typename U>
ostream &operator<<(ostream &s, const set<U> &x) {
  s << "{";
  bool was = false;
  for (auto it : x) {
    if (was) {
      s << ", ";
    }
    was = true;
    s << it;
  }
  s << "}";
  return s;
}
int main() {
  srand(time(NULL));
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> a[i].second;
  }
  map<int, vector<pair<int, int>>> b;
  for (int i = 0; i < n; i++) {
    b[a[i].first].push_back({a[i].second, i});
  }
  sort(a.begin(), a.end());
  long long ans = 0;
  int ptr = 0;
  set<pair<int, int>> was;
  long long was_sum = 0;
  while (true) {
    if (was.empty()) {
      auto it = b.lower_bound(ptr);
      if (it == b.end()) {
        break;
      }
      ptr = it->first;
    }
    if (b.count(ptr)) {
      for (auto kv : b[ptr]) {
        was.insert(kv);
        was_sum += kv.first;
      }
    }
    assert(!was.empty());
    auto val = *was.rbegin();
    was.erase(val);
    was_sum -= val.first;
    ans += was_sum;
    ++ptr;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 double pi = acos(-1.0);
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
bool chmin(T &x, T y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
bool chmax(T &x, T y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
template <typename U, typename V>
ostream &operator<<(ostream &s, const pair<U, V> &x) {
  s << "(" << x.first << ", " << x.second << ")";
  return s;
}
template <typename U>
ostream &operator<<(ostream &s, const vector<U> &x) {
  s << "[";
  bool was = false;
  for (auto it : x) {
    if (was) {
      s << ", ";
    }
    was = true;
    s << it;
  }
  s << "]";
  return s;
}
template <typename U>
ostream &operator<<(ostream &s, const set<U> &x) {
  s << "{";
  bool was = false;
  for (auto it : x) {
    if (was) {
      s << ", ";
    }
    was = true;
    s << it;
  }
  s << "}";
  return s;
}
int main() {
  srand(time(NULL));
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> a[i].second;
  }
  map<int, vector<pair<int, int>>> b;
  for (int i = 0; i < n; i++) {
    b[a[i].first].push_back({a[i].second, i});
  }
  sort(a.begin(), a.end());
  long long ans = 0;
  int ptr = 0;
  set<pair<int, int>> was;
  long long was_sum = 0;
  while (true) {
    if (was.empty()) {
      auto it = b.lower_bound(ptr);
      if (it == b.end()) {
        break;
      }
      ptr = it->first;
    }
    if (b.count(ptr)) {
      for (auto kv : b[ptr]) {
        was.insert(kv);
        was_sum += kv.first;
      }
    }
    assert(!was.empty());
    auto val = *was.rbegin();
    was.erase(val);
    was_sum -= val.first;
    ans += was_sum;
    ++ptr;
  }
  cout << ans << endl;
  return 0;
}

```