#include <bits/stdc++.h>
using namespace std;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
T sqr(T a) {
  return a * a;
}
template <class T>
T sgn(T a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
const double pi = acos(-1.0);
int n;
int m;
int main() {
  cin >> n;
  vector<pair<int, int> > a(n);
  vector<int> t(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[i] = make_pair(x, i);
  }
  for (int i = 0; i < n; i++) cin >> t[i];
  sort((a).begin(), (a).end());
  long long s = 0, ans = 0;
  int cnt = 0;
  set<pair<int, int> > q;
  for (int i = 0; i < n || ((int)(q).size()) > 0;) {
    if (i < n) {
      int id = a[i].second, val = a[i].first;
      if (((int)(q).size()) == 0 || (((int)(q).size()) > 0 && val <= cnt)) {
        q.insert(make_pair(t[id], id));
        s += t[id];
        while (i + 1 < n && a[i].first == a[i + 1].first) {
          i++;
          int id = a[i].second;
          q.insert(make_pair(t[id], id));
          s += t[id];
        }
        cnt = max(val, cnt);
        i++;
      }
    }
    pair<int, int> cur = *(--q.end());
    q.erase(--q.end());
    s -= cur.first;
    ans += s;
    cnt++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
T sqr(T a) {
  return a * a;
}
template <class T>
T sgn(T a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
const double pi = acos(-1.0);
int n;
int m;
int main() {
  cin >> n;
  vector<pair<int, int> > a(n);
  vector<int> t(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[i] = make_pair(x, i);
  }
  for (int i = 0; i < n; i++) cin >> t[i];
  sort((a).begin(), (a).end());
  long long s = 0, ans = 0;
  int cnt = 0;
  set<pair<int, int> > q;
  for (int i = 0; i < n || ((int)(q).size()) > 0;) {
    if (i < n) {
      int id = a[i].second, val = a[i].first;
      if (((int)(q).size()) == 0 || (((int)(q).size()) > 0 && val <= cnt)) {
        q.insert(make_pair(t[id], id));
        s += t[id];
        while (i + 1 < n && a[i].first == a[i + 1].first) {
          i++;
          int id = a[i].second;
          q.insert(make_pair(t[id], id));
          s += t[id];
        }
        cnt = max(val, cnt);
        i++;
      }
    }
    pair<int, int> cur = *(--q.end());
    q.erase(--q.end());
    s -= cur.first;
    ans += s;
    cnt++;
  }
  cout << ans << endl;
  return 0;
}

```