#include <bits/stdc++.h>
using namespace std;
inline void solve() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  long long sum = 0;
  map<int, vector<int>> st;
  for (int i = 0; i < n; i++) {
    st[a[i]].push_back(b[i]);
  }
  multiset<int, greater<int>> mins;
  long long cursum = 0;
  auto it = st.begin();
  while (it != st.end()) {
    int val = it->first;
    vector<int> nums = it->second;
    if ((int)(mins).size() > 0 || (int)(nums).size() > 1) {
      for (int i : nums) {
        mins.insert(i);
        cursum += i;
      }
      cursum -= *mins.begin();
      mins.erase(mins.begin());
      if (st.find(val + 1) == st.end()) {
        st[val + 1] = vector<int>();
      }
    }
    sum += cursum;
    it++;
  }
  cout << sum << '\n';
}
int32_t main() {
  clock_t startTime;
  startTime = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << setprecision(10);
  int t = 1;
  while (t--) {
    solve();
  }
  fflush(stdin);
  fflush(stdout);
  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;
inline void solve() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  long long sum = 0;
  map<int, vector<int>> st;
  for (int i = 0; i < n; i++) {
    st[a[i]].push_back(b[i]);
  }
  multiset<int, greater<int>> mins;
  long long cursum = 0;
  auto it = st.begin();
  while (it != st.end()) {
    int val = it->first;
    vector<int> nums = it->second;
    if ((int)(mins).size() > 0 || (int)(nums).size() > 1) {
      for (int i : nums) {
        mins.insert(i);
        cursum += i;
      }
      cursum -= *mins.begin();
      mins.erase(mins.begin());
      if (st.find(val + 1) == st.end()) {
        st[val + 1] = vector<int>();
      }
    }
    sum += cursum;
    it++;
  }
  cout << sum << '\n';
}
int32_t main() {
  clock_t startTime;
  startTime = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << setprecision(10);
  int t = 1;
  while (t--) {
    solve();
  }
  fflush(stdin);
  fflush(stdout);
  return 0;
}

```