#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int MAX = 1000;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, m;
struct re {
  int a, t;
} a[maxn];
priority_queue<int> q;
bool cmp(re a, re b) {
  if (a.a == b.a) return a.t < b.t;
  return a.a < b.a;
}
signed main() {
  int n;
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i].a);
  }
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i].t);
  }
  sort(a + 1, a + 1 + n, cmp);
  long long ans = 0;
  long long sum;
  int lay = 0;
  a[n + 1].a = 2e9;
  for (int i = (1); i <= (n + 1); ++i) {
    while (a[i].a > lay) {
      if (q.size() <= 1) {
        while (!q.empty()) q.pop();
        sum = 0;
        lay = a[i].a;
        break;
      }
      lay++;
      sum -= 1ll * q.top();
      ans += sum;
      q.pop();
    }
    q.push(a[i].t);
    sum += a[i].t;
  }
  cout << (ans) << endl;
}

### 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 = 2e5 + 10;
const int MAX = 1000;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, m;
struct re {
  int a, t;
} a[maxn];
priority_queue<int> q;
bool cmp(re a, re b) {
  if (a.a == b.a) return a.t < b.t;
  return a.a < b.a;
}
signed main() {
  int n;
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i].a);
  }
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i].t);
  }
  sort(a + 1, a + 1 + n, cmp);
  long long ans = 0;
  long long sum;
  int lay = 0;
  a[n + 1].a = 2e9;
  for (int i = (1); i <= (n + 1); ++i) {
    while (a[i].a > lay) {
      if (q.size() <= 1) {
        while (!q.empty()) q.pop();
        sum = 0;
        lay = a[i].a;
        break;
      }
      lay++;
      sum -= 1ll * q.top();
      ans += sum;
      q.pop();
    }
    q.push(a[i].t);
    sum += a[i].t;
  }
  cout << (ans) << endl;
}

```