#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long powmod(long long a, long long b, long long p) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long temp = powmod(a, b / 2, p);
  temp *= temp;
  temp %= p;
  if (b % 2 == 0) return temp;
  temp = temp * a;
  temp %= p;
  return temp;
}
void solve() {
  int n;
  cin >> n;
  int a[n], b[n];
  for (__typeof((n)) i = (0); i < (n); i++) cin >> a[i];
  for (__typeof((n)) i = (0); i < (n); i++) cin >> b[i];
  vector<pair<int, int> > v;
  for (__typeof((n)) i = (0); i < (n); i++) v.push_back({a[i], b[i]});
  sort((v).begin(), (v).end());
  priority_queue<int> pq;
  long long ans = 0, sum = 0, val = 0;
  for (__typeof((n + 1)) i = (0); i < (n + 1); i++) {
    while (!pq.empty() && (i == n || v[i].first > val)) {
      sum -= pq.top();
      pq.pop();
      ans += sum;
      val++;
    }
    if (i == n) break;
    val = v[i].first;
    sum += v[i].second;
    pq.push(v[i].second);
  }
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long powmod(long long a, long long b, long long p) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long temp = powmod(a, b / 2, p);
  temp *= temp;
  temp %= p;
  if (b % 2 == 0) return temp;
  temp = temp * a;
  temp %= p;
  return temp;
}
void solve() {
  int n;
  cin >> n;
  int a[n], b[n];
  for (__typeof((n)) i = (0); i < (n); i++) cin >> a[i];
  for (__typeof((n)) i = (0); i < (n); i++) cin >> b[i];
  vector<pair<int, int> > v;
  for (__typeof((n)) i = (0); i < (n); i++) v.push_back({a[i], b[i]});
  sort((v).begin(), (v).end());
  priority_queue<int> pq;
  long long ans = 0, sum = 0, val = 0;
  for (__typeof((n + 1)) i = (0); i < (n + 1); i++) {
    while (!pq.empty() && (i == n || v[i].first > val)) {
      sum -= pq.top();
      pq.pop();
      ans += sum;
      val++;
    }
    if (i == n) break;
    val = v[i].first;
    sum += v[i].second;
    pq.push(v[i].second);
  }
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```