#include <bits/stdc++.h>
using namespace std;
const int M = 222000;
const int inf = 0x3f3f3f3f;
int n, m, k;
priority_queue<int> qq;
int a[M];
int t[M];
pair<int, int> v[M];
long long solve() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (n); ++i) cin >> t[i];
  for (int i = (1); i <= (n); ++i) v[i] = {a[i], t[i]};
  sort(v + 1, v + 1 + n);
  int pre = 0;
  long long sum = 0;
  long long res = 0;
  v[n + 1] = {inf, 0};
  for (int i = (1); i <= (n + 1); ++i) {
    if (v[i].first > pre) {
      while (pre < v[i].first && !qq.empty()) {
        int tp = qq.top();
        qq.pop();
        sum -= tp;
        res += sum;
        ++pre;
      }
      res += sum * (v[i].first - pre);
      pre = v[i].first;
    }
    qq.push(v[i].second);
    sum += v[i].second;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << solve() << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 M = 222000;
const int inf = 0x3f3f3f3f;
int n, m, k;
priority_queue<int> qq;
int a[M];
int t[M];
pair<int, int> v[M];
long long solve() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (n); ++i) cin >> t[i];
  for (int i = (1); i <= (n); ++i) v[i] = {a[i], t[i]};
  sort(v + 1, v + 1 + n);
  int pre = 0;
  long long sum = 0;
  long long res = 0;
  v[n + 1] = {inf, 0};
  for (int i = (1); i <= (n + 1); ++i) {
    if (v[i].first > pre) {
      while (pre < v[i].first && !qq.empty()) {
        int tp = qq.top();
        qq.pop();
        sum -= tp;
        res += sum;
        ++pre;
      }
      res += sum * (v[i].first - pre);
      pre = v[i].first;
    }
    qq.push(v[i].second);
    sum += v[i].second;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << solve() << endl;
  return 0;
}

```