#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1E9 + 7, INF = 2E18 + 5;
const double PI = 2 * acos(0.0);
const long double EPS = 1.0E-14;
pair<long long, long long> pt[200005];
bool comp(const pair<long long, long long> &A,
          const pair<long long, long long> &B) {
  if (A.first == B.first)
    return A.second < B.second;
  else
    return A.first < B.first;
}
priority_queue<long long> pq;
int main() {
  long long n, ans = 0, cur = 0, sum = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &pt[i].first);
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &pt[i].second);
  }
  sort(pt, pt + n, comp);
  for (int i = 0; i < n; i++) {
    if (cur == pt[i].first) {
      pq.push(pt[i].second);
      sum += pt[i].second;
    } else {
      if (!pq.empty()) {
        sum -= pq.top();
        pq.pop();
        ans += sum;
        cur++;
        i--;
      } else {
        sum += pt[i].second;
        cur = pt[i].first;
        pq.push(pt[i].second);
      }
    }
  }
  while (!pq.empty()) {
    sum -= pq.top();
    pq.pop();
    ans += sum;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const long long MOD = 1E9 + 7, INF = 2E18 + 5;
const double PI = 2 * acos(0.0);
const long double EPS = 1.0E-14;
pair<long long, long long> pt[200005];
bool comp(const pair<long long, long long> &A,
          const pair<long long, long long> &B) {
  if (A.first == B.first)
    return A.second < B.second;
  else
    return A.first < B.first;
}
priority_queue<long long> pq;
int main() {
  long long n, ans = 0, cur = 0, sum = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &pt[i].first);
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &pt[i].second);
  }
  sort(pt, pt + n, comp);
  for (int i = 0; i < n; i++) {
    if (cur == pt[i].first) {
      pq.push(pt[i].second);
      sum += pt[i].second;
    } else {
      if (!pq.empty()) {
        sum -= pq.top();
        pq.pop();
        ans += sum;
        cur++;
        i--;
      } else {
        sum += pt[i].second;
        cur = pt[i].first;
        pq.push(pt[i].second);
      }
    }
  }
  while (!pq.empty()) {
    sum -= pq.top();
    pq.pop();
    ans += sum;
  }
  printf("%lld\n", ans);
  return 0;
}

```