#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 5;
struct Node {
  int x, y;
};
Node a[200005];
int n;
bool cmp(Node x, Node y) { return x.x < y.x; }
signed main() {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].x;
  for (int i = 1; i <= n; i++) cin >> a[i].y;
  sort(a + 1, a + n + 1, cmp);
  priority_queue<int> q;
  long long ind = 0, cnt = 0, ans = 0, tmp = 0;
  while (!q.empty() || ind < n) {
    cnt++;
    if (q.empty()) {
      cnt = a[ind + 1].x;
      while (cnt == a[ind + 1].x && ind < n) {
        ind++;
        tmp += a[ind].y;
        q.push(a[ind].y);
      }
    } else {
      while (cnt == a[ind + 1].x && ind < n) {
        ind++;
        tmp += a[ind].y;
        q.push(a[ind].y);
      }
    }
    tmp -= q.top();
    q.pop();
    ans += tmp;
  }
  cout << ans << 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 mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 5;
struct Node {
  int x, y;
};
Node a[200005];
int n;
bool cmp(Node x, Node y) { return x.x < y.x; }
signed main() {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].x;
  for (int i = 1; i <= n; i++) cin >> a[i].y;
  sort(a + 1, a + n + 1, cmp);
  priority_queue<int> q;
  long long ind = 0, cnt = 0, ans = 0, tmp = 0;
  while (!q.empty() || ind < n) {
    cnt++;
    if (q.empty()) {
      cnt = a[ind + 1].x;
      while (cnt == a[ind + 1].x && ind < n) {
        ind++;
        tmp += a[ind].y;
        q.push(a[ind].y);
      }
    } else {
      while (cnt == a[ind + 1].x && ind < n) {
        ind++;
        tmp += a[ind].y;
        q.push(a[ind].y);
      }
    }
    tmp -= q.top();
    q.pop();
    ans += tmp;
  }
  cout << ans << endl;
  return 0;
}

```