#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 2e5;
struct mode {
  int a, t;
} b[N + 5];
int n;
map<int, int> mp;
long long ans;
int find(int x, long long pay) {
  if (mp[x]) {
    ans += (mp[x] - x) * pay;
    return mp[x] = find(mp[x], pay);
  } else {
    mp[x] = x + 1;
    return x + 1;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i].a;
  for (int i = 1; i <= n; i++) cin >> b[i].t;
  sort(b + 1, b + 1 + n,
       [](const mode &m1, const mode &m2) -> bool { return m1.t > m2.t; });
  for (int i = 1; i <= n; i++) {
    if (!mp[b[i].a])
      mp[b[i].a] = b[i].a + 1;
    else
      find(b[i].a, b[i].t);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int INF = 0x3f3f3f3f;
const int N = 2e5;
struct mode {
  int a, t;
} b[N + 5];
int n;
map<int, int> mp;
long long ans;
int find(int x, long long pay) {
  if (mp[x]) {
    ans += (mp[x] - x) * pay;
    return mp[x] = find(mp[x], pay);
  } else {
    mp[x] = x + 1;
    return x + 1;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i].a;
  for (int i = 1; i <= n; i++) cin >> b[i].t;
  sort(b + 1, b + 1 + n,
       [](const mode &m1, const mode &m2) -> bool { return m1.t > m2.t; });
  for (int i = 1; i <= n; i++) {
    if (!mp[b[i].a])
      mp[b[i].a] = b[i].a + 1;
    else
      find(b[i].a, b[i].t);
  }
  cout << ans << endl;
  return 0;
}

```