#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
struct node {
  int id, w;
  bool friend operator<(const node& a, const node& b) { return a.id < b.id; }
};
int a[N], b[N];
void solve() {
  int n;
  scanf("%d", &n);
  vector<node> vn;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", b + i);
  }
  for (int i = 1; i <= n; ++i) {
    vn.push_back(node{a[i], b[i]});
  }
  sort(vn.begin(), vn.end());
  priority_queue<int, vector<int>, less<int> > que;
  int pre_id, len, tot;
  long long sum, ans;
  pre_id = -1;
  len = tot = sum = ans = 0;
  while (1) {
    int now_id = vn[len].id;
    if (pre_id != -1) {
      int d = now_id - pre_id;
      while (!que.empty() && d) {
        sum -= que.top();
        que.pop();
        ans += sum;
        tot++;
        d--;
      }
    }
    for (int i = len; i <= n; ++i) {
      if (i == n) {
        len = n;
        break;
      } else if (vn[i].id != now_id) {
        len = i;
        break;
      } else {
        que.push(vn[i].w);
        sum += vn[i].w;
      }
    }
    if (len == n) break;
    pre_id = now_id;
  }
  while (1) {
    if (tot == n) break;
    sum -= que.top();
    que.pop();
    ans += sum;
    tot++;
  }
  printf("%lld\n", ans);
}
int main() {
  solve();
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 N = 2e5 + 10;
struct node {
  int id, w;
  bool friend operator<(const node& a, const node& b) { return a.id < b.id; }
};
int a[N], b[N];
void solve() {
  int n;
  scanf("%d", &n);
  vector<node> vn;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", b + i);
  }
  for (int i = 1; i <= n; ++i) {
    vn.push_back(node{a[i], b[i]});
  }
  sort(vn.begin(), vn.end());
  priority_queue<int, vector<int>, less<int> > que;
  int pre_id, len, tot;
  long long sum, ans;
  pre_id = -1;
  len = tot = sum = ans = 0;
  while (1) {
    int now_id = vn[len].id;
    if (pre_id != -1) {
      int d = now_id - pre_id;
      while (!que.empty() && d) {
        sum -= que.top();
        que.pop();
        ans += sum;
        tot++;
        d--;
      }
    }
    for (int i = len; i <= n; ++i) {
      if (i == n) {
        len = n;
        break;
      } else if (vn[i].id != now_id) {
        len = i;
        break;
      } else {
        que.push(vn[i].w);
        sum += vn[i].w;
      }
    }
    if (len == n) break;
    pre_id = now_id;
  }
  while (1) {
    if (tot == n) break;
    sum -= que.top();
    que.pop();
    ans += sum;
    tot++;
  }
  printf("%lld\n", ans);
}
int main() {
  solve();
  return 0;
}

```