#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  cin >> x;
}
template <typename T, typename T0>
void read(T &x, T0 &y) {
  cin >> x >> y;
}
template <typename T, typename T0, typename T1>
void read(T &x, T0 &y, T1 &z) {
  cin >> x >> y >> z;
}
template <typename T, typename T0, typename T1, typename T2>
void read(T &x, T0 &y, T1 &z, T2 &w) {
  cin >> x >> y >> z >> w;
}
template <typename T, typename T0>
void read(pair<T, T0> &p) {
  cin >> p.first >> p.second;
}
template <typename T>
void read(vector<T> &oneD, long long int n) {
  for (long long int(i) = 0; (i) < (n); ++(i)) {
    long long int x;
    read(x);
    oneD.push_back(x);
  }
}
template <typename T>
void read(T oneD[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    read(oneD[i]);
  }
}
template <typename T>
void write(T &x) {
  cout << x << " ";
}
template <typename T>
void wrtline(T &x) {
  cout << x << "\n";
}
template <typename T, typename T0>
void write(T &x, T0 &y) {
  cout << x << " " << y << "\n";
}
template <typename T, typename T0, typename T1>
void write(T &x, T0 &y, T1 &z) {
  cout << x << " " << y << " " << z << "\n";
}
template <typename T, typename T0, typename T1, typename T2>
void write(T &x, T0 &y, T1 &z, T2 &w) {
  cout << x << " " << y << " " << z << " " << w << "\n";
}
template <typename T, typename T0>
void write(pair<T, T0> &p) {
  write(p.first);
  write(p.second);
  cout << "\n";
}
template <typename T>
void write(vector<T> &oneD, long long int n) {
  for (long long int i = 0; i < n; i++) {
    cout << oneD[i] << " ";
  }
  cout << "\n";
}
template <typename T>
void write(T oneD[], int n) {
  for (long long int i = 0; i < n; i++) {
    write(oneD[i]);
  }
  cout << "\n";
}
template <typename T, typename T0>
void write(map<T, T0> &mpp) {
  for (auto it : mpp) {
    write(it.first);
    cout << ": ";
    write(it.second);
    cout << "\n";
  }
  cout << "\n";
}
void write(vector<vector<long long int>> rc, long long int n, long long int m) {
  for (long long int(i) = 0; (i) < (n); ++(i)) {
    for (long long int(j) = 0; (j) < (m); ++(j)) {
      cout << rc[i][j] << " ";
    }
    cout << "\n";
  }
}
void solve() {
  long long int n;
  read(n);
  vector<pair<long long int, long long int>> p(n);
  for (long long int(i) = 0; (i) < (n); ++(i)) read(p[i].first);
  for (long long int(i) = 0; (i) < (n); ++(i)) read(p[i].second);
  sort((p).begin(), (p).end());
  priority_queue<long long int> q;
  long long int sum = 0, ct = 0, val = 0;
  for (long long int(i) = 0; (i) < (n + 1); ++(i)) {
    while (q.size() and (i == n or p[i].first > val)) {
      sum -= q.top();
      q.pop();
      ct += sum;
      val++;
    }
    val = p[i].first;
    sum += p[i].second;
    q.push(p[i].second);
  }
  wrtline(ct);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

### Prompt

Create a solution in cpp for 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;
template <typename T>
void read(T &x) {
  cin >> x;
}
template <typename T, typename T0>
void read(T &x, T0 &y) {
  cin >> x >> y;
}
template <typename T, typename T0, typename T1>
void read(T &x, T0 &y, T1 &z) {
  cin >> x >> y >> z;
}
template <typename T, typename T0, typename T1, typename T2>
void read(T &x, T0 &y, T1 &z, T2 &w) {
  cin >> x >> y >> z >> w;
}
template <typename T, typename T0>
void read(pair<T, T0> &p) {
  cin >> p.first >> p.second;
}
template <typename T>
void read(vector<T> &oneD, long long int n) {
  for (long long int(i) = 0; (i) < (n); ++(i)) {
    long long int x;
    read(x);
    oneD.push_back(x);
  }
}
template <typename T>
void read(T oneD[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    read(oneD[i]);
  }
}
template <typename T>
void write(T &x) {
  cout << x << " ";
}
template <typename T>
void wrtline(T &x) {
  cout << x << "\n";
}
template <typename T, typename T0>
void write(T &x, T0 &y) {
  cout << x << " " << y << "\n";
}
template <typename T, typename T0, typename T1>
void write(T &x, T0 &y, T1 &z) {
  cout << x << " " << y << " " << z << "\n";
}
template <typename T, typename T0, typename T1, typename T2>
void write(T &x, T0 &y, T1 &z, T2 &w) {
  cout << x << " " << y << " " << z << " " << w << "\n";
}
template <typename T, typename T0>
void write(pair<T, T0> &p) {
  write(p.first);
  write(p.second);
  cout << "\n";
}
template <typename T>
void write(vector<T> &oneD, long long int n) {
  for (long long int i = 0; i < n; i++) {
    cout << oneD[i] << " ";
  }
  cout << "\n";
}
template <typename T>
void write(T oneD[], int n) {
  for (long long int i = 0; i < n; i++) {
    write(oneD[i]);
  }
  cout << "\n";
}
template <typename T, typename T0>
void write(map<T, T0> &mpp) {
  for (auto it : mpp) {
    write(it.first);
    cout << ": ";
    write(it.second);
    cout << "\n";
  }
  cout << "\n";
}
void write(vector<vector<long long int>> rc, long long int n, long long int m) {
  for (long long int(i) = 0; (i) < (n); ++(i)) {
    for (long long int(j) = 0; (j) < (m); ++(j)) {
      cout << rc[i][j] << " ";
    }
    cout << "\n";
  }
}
void solve() {
  long long int n;
  read(n);
  vector<pair<long long int, long long int>> p(n);
  for (long long int(i) = 0; (i) < (n); ++(i)) read(p[i].first);
  for (long long int(i) = 0; (i) < (n); ++(i)) read(p[i].second);
  sort((p).begin(), (p).end());
  priority_queue<long long int> q;
  long long int sum = 0, ct = 0, val = 0;
  for (long long int(i) = 0; (i) < (n + 1); ++(i)) {
    while (q.size() and (i == n or p[i].first > val)) {
      sum -= q.top();
      q.pop();
      ct += sum;
      val++;
    }
    val = p[i].first;
    sum += p[i].second;
    q.push(p[i].second);
  }
  wrtline(ct);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

```