#include <bits/stdc++.h>
using namespace std;
struct smoke {
  int a;
  int time;
} c[200010];
int cmp(smoke a, smoke b) {
  if (a.a == b.a) return a.time < b.time;
  return a.a < b.a;
}
int main() {
  int n;
  int i;
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%d", &c[i].a);
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &c[i].time);
  }
  sort(c + 1, c + n + 1, cmp);
  int now = 0;
  long long sum = 0;
  long long all = 0;
  map<int, int> mp;
  priority_queue<int> q;
  for (i = 1; i <= n; i++) {
    if (c[i].a == c[i - 1].a) {
      if (c[i].time < mp[c[i].a]) {
        q.push(c[i].time);
        sum += c[i].time;
      } else {
        q.push(mp[c[i].a]);
        sum += mp[c[i].a];
        mp[c[i].a] = c[i].time;
      }
      now = c[i].a;
    } else {
      while (now < c[i].a) {
        if (sum == 0) break;
        all += sum;
        now++;
        if (now < c[i].a) {
          sum -= q.top();
          q.pop();
        }
      }
      if (!q.empty()) {
        if (c[i].time < q.top()) {
          sum -= q.top();
          sum += c[i].time;
          mp[c[i].a] = q.top();
          q.pop();
          q.push(c[i].time);
        } else
          mp[c[i].a] = c[i].time;
      } else
        mp[c[i].a] = c[i].time;
    }
  }
  while (!q.empty()) {
    all += sum;
    sum -= q.top();
    q.pop();
  }
  cout << all << endl;
}

### 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;
struct smoke {
  int a;
  int time;
} c[200010];
int cmp(smoke a, smoke b) {
  if (a.a == b.a) return a.time < b.time;
  return a.a < b.a;
}
int main() {
  int n;
  int i;
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%d", &c[i].a);
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &c[i].time);
  }
  sort(c + 1, c + n + 1, cmp);
  int now = 0;
  long long sum = 0;
  long long all = 0;
  map<int, int> mp;
  priority_queue<int> q;
  for (i = 1; i <= n; i++) {
    if (c[i].a == c[i - 1].a) {
      if (c[i].time < mp[c[i].a]) {
        q.push(c[i].time);
        sum += c[i].time;
      } else {
        q.push(mp[c[i].a]);
        sum += mp[c[i].a];
        mp[c[i].a] = c[i].time;
      }
      now = c[i].a;
    } else {
      while (now < c[i].a) {
        if (sum == 0) break;
        all += sum;
        now++;
        if (now < c[i].a) {
          sum -= q.top();
          q.pop();
        }
      }
      if (!q.empty()) {
        if (c[i].time < q.top()) {
          sum -= q.top();
          sum += c[i].time;
          mp[c[i].a] = q.top();
          q.pop();
          q.push(c[i].time);
        } else
          mp[c[i].a] = c[i].time;
      } else
        mp[c[i].a] = c[i].time;
    }
  }
  while (!q.empty()) {
    all += sum;
    sum -= q.top();
    q.pop();
  }
  cout << all << endl;
}

```