#include <bits/stdc++.h>
using namespace std;
inline void pisz(int n) { printf("%d\n", n); }
const int fx[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int fxx[8][2] = {{0, 1}, {0, -1}, {1, 0},  {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int i = 0; i < ((int)((t).size())); i++) s << t[i] << " ";
  return s;
}
int main() {
  int n;
  scanf("%d", &n);
  int arr[n];
  map<int, int> occurance;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    occurance[arr[i]]++;
  }
  vector<pair<int, int> > withTime;
  map<int, int> time;
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    withTime.push_back(make_pair(arr[i], t));
  }
  sort(withTime.begin(), withTime.end());
  reverse(withTime.begin(), withTime.end());
  priority_queue<int> timeTrack;
  int idx;
  long long curVal = 0;
  long long ans = 0;
  while (!withTime.empty() || !timeTrack.empty()) {
    if (timeTrack.empty()) {
      idx = withTime.back().first;
    }
    while (!withTime.empty() && withTime.back().first == idx) {
      curVal += withTime.back().second;
      timeTrack.push(withTime.back().second);
      withTime.pop_back();
    }
    if (!timeTrack.empty()) {
      curVal -= timeTrack.top();
      timeTrack.pop();
    }
    idx++;
    ans += curVal;
  }
  cout << ans << endl;
}

### 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;
inline void pisz(int n) { printf("%d\n", n); }
const int fx[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int fxx[8][2] = {{0, 1}, {0, -1}, {1, 0},  {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int i = 0; i < ((int)((t).size())); i++) s << t[i] << " ";
  return s;
}
int main() {
  int n;
  scanf("%d", &n);
  int arr[n];
  map<int, int> occurance;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    occurance[arr[i]]++;
  }
  vector<pair<int, int> > withTime;
  map<int, int> time;
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    withTime.push_back(make_pair(arr[i], t));
  }
  sort(withTime.begin(), withTime.end());
  reverse(withTime.begin(), withTime.end());
  priority_queue<int> timeTrack;
  int idx;
  long long curVal = 0;
  long long ans = 0;
  while (!withTime.empty() || !timeTrack.empty()) {
    if (timeTrack.empty()) {
      idx = withTime.back().first;
    }
    while (!withTime.empty() && withTime.back().first == idx) {
      curVal += withTime.back().second;
      timeTrack.push(withTime.back().second);
      withTime.pop_back();
    }
    if (!timeTrack.empty()) {
      curVal -= timeTrack.top();
      timeTrack.pop();
    }
    idx++;
    ans += curVal;
  }
  cout << ans << endl;
}

```