#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
const int inf = 1e9 + 5;
int n;
pair<long long, long long> niz[maxn];
priority_queue<long long> pq;
int main() {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> niz[i].first;
  for (int i = 1; i <= n; i++) cin >> niz[i].second;
  sort(niz + 1, niz + n + 1);
  long long rez = 0;
  long long zbir = 0;
  for (int i = 1; i <= n; i++) {
    long long x = niz[i].first;
    long long y = niz[i].second;
    pq.push(y);
    zbir += y;
    while (i + 1 <= n && niz[i + 1].first == niz[i].first) {
      i++;
      zbir += niz[i].second;
      pq.push(niz[i].second);
    }
    long long pom = niz[i].first;
    while (!pq.empty() && pom != niz[i + 1].first) {
      zbir -= pq.top();
      rez += zbir;
      pq.pop();
      pom++;
    }
  }
  while (!pq.empty()) {
    zbir -= pq.top();
    rez += zbir;
    pq.pop();
  }
  cout << rez << endl;
  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;
const int maxn = 200005;
const int inf = 1e9 + 5;
int n;
pair<long long, long long> niz[maxn];
priority_queue<long long> pq;
int main() {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> niz[i].first;
  for (int i = 1; i <= n; i++) cin >> niz[i].second;
  sort(niz + 1, niz + n + 1);
  long long rez = 0;
  long long zbir = 0;
  for (int i = 1; i <= n; i++) {
    long long x = niz[i].first;
    long long y = niz[i].second;
    pq.push(y);
    zbir += y;
    while (i + 1 <= n && niz[i + 1].first == niz[i].first) {
      i++;
      zbir += niz[i].second;
      pq.push(niz[i].second);
    }
    long long pom = niz[i].first;
    while (!pq.empty() && pom != niz[i + 1].first) {
      zbir -= pq.top();
      rez += zbir;
      pq.pop();
      pom++;
    }
  }
  while (!pq.empty()) {
    zbir -= pq.top();
    rez += zbir;
    pq.pop();
  }
  cout << rez << endl;
  return 0;
}

```